Engineering Mechanics: Dynamics in Si Units [14 ed.]
 1292088729, 9781292088723

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ENGINEERING MECHANICS

DYNAMICS FourTEENTH EDITIoN IN sI uNITs

r. C. HIBBELEr SI Conversion by

Kai Beng Yap

Hoboken Boston Columbus San Francisco New York Indianapolis London  Toronto Sydney Singapore Tokyo Montreal Dubai Madrid Hong Kong Mexico City Munich Paris Amsterdam Cape Town

Vice President and Editorial Director, ECS: Marcia Horton Senior Editor: Norrin Dias Acquisitions Editor, Global Editions: Murchana Borthakur Editorial Assistant: Michelle Bayman Program and Project Management Team Lead: Scott Disanno Program Manager: Sandra L. Rodriguez Project Manager: Rose Kernan Project Editor, Global Editions: Donald Villamero Art Editor: Gregory Dulles Senior Digital Producer: Felipe Gonzalez Operations Specialist: Maura Zaldivar-Garcia Senior Production Manufacturing Controller, Global Editions: Trudy Kimber Media Production Manager, Global Editions: Vikram Kumar Product Marketing Manager: Bram Van Kempen Field Marketing Manager: Demetrius Hall Marketing Assistant: Jon Bryant Cover Image: Italianvideophotoagency/Shutterstock Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsonglobaleditions.com © 2017 by R. C. Hibbeler The rights of R. C. Hibbeler to be identified as the author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. Authorized adaptation from the United States edition, entitled Engineering Mechanics: Dynamics, Fourteenth Edition, ISBN 978-0-13-391538-9, by R. C. Hibbeler, published by Pearson Education, Inc., publishing as Pearson Prentice Hall © 2016. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. Many of the designations by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook appear on the appropriate page within the text. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. 10 9 8 7 6 5 4 3 2 1 ISBN 10: 1-292-08872-9 ISBN 13: 978-1-292-08872-3 Printed and bound by L.E.G.O. S.p.A., Italy

To the student With the hope that this work will stimulate an interest in Engineering Mechanics and provide an acceptable guide to its understanding.

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PREFACE

The main purpose of this book is to provide the student with a clear and thorough presentation of the theory and application of engineering mechanics. To achieve this objective, this work has been shaped by the comments and suggestions of hundreds of reviewers in the teaching profession, as well as many of the author’s students.

New to this Edition Preliminary Problems. This new feature can be found throughout the text, and is given just before the Fundamental Problems. The intent here is to test the student’s conceptual understanding of the theory. Normally the solutions require little or no calculation, and as such, these problems provide a basic understanding of the concepts before they are applied numerically. All the solutions are given in the back of the text.

Expanded Important Points Sections. Summaries have been added which reinforces the reading material and highlights the important definitions and concepts of the sections.

Re-writing of Text Material. Further clarification of concepts has been included in this edition, and important definitions are now in boldface throughout the text to highlight their importance.

End-of-the-Chapter Review Problems. All the review problems now have solutions given in the back, so that students can check their work when studying for exams, and reviewing their skills when the chapter is finished. New Photos. The relevance of knowing the subject matter is reflected by the

real-world applications depicted in the over 30 new or updated photos placed throughout the book. These photos generally are used to explain how the relevant principles apply to real-world situations and how materials behave under load.

New Problems. There are approximately 30% new problems that have been

added to this edition, which involve applications to many different fields of engineering. v

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Preface

Hallmark Features Besides the new features mentioned above, other outstanding features that define the contents of the text include the following.

Organization and Approach. Each chapter is organized into well-defined

sections that contain an explanation of specific topics, illustrative example problems, and a set of homework problems. The topics within each section are placed into subgroups defined by boldface titles. The purpose of this is to present a structured method for introducing each new definition or concept and to make the book convenient for later reference and review.

Chapter Contents. Each chapter begins with an illustration demonstrating a broad-range application of the material within the chapter. A bulleted list of the chapter contents is provided to give a general overview of the material that will be covered.

Emphasis on Free-Body Diagrams. Drawing a free-body diagram is

particularly important when solving problems, and for this reason this step is strongly emphasized throughout the book. In particular, special sections and examples are devoted to show how to draw free-body diagrams. Specific homework problems have also been added to develop this practice.

Procedures for Analysis. A general procedure for analyzing any mechanical problem is presented at the end of the first chapter. Then this procedure is customized to relate to specific types of problems that are covered throughout the book. This unique feature provides the student with a logical and orderly method to follow when applying the theory. The example problems are solved using this outlined method in order to clarify its numerical application. Realize, however, that once the relevant principles have been mastered and enough confidence and judgment have been obtained, the student can then develop his or her own procedures for solving problems. Important Points. This feature provides a review or summary of the most

important concepts in a section and highlights the most significant points that should be realized when applying the theory to solve problems.

Fundamental Problems. These problem sets are selectively located just after most of the example problems. They provide students with simple applications of the concepts, and therefore, the chance to develop their problem-solving skills before attempting to solve any of the standard problems that follow. In addition, they can be used for preparing for exams. Conceptual Understanding. Through the use of photographs placed throughout the book, theory is applied in a simplified way in order to illustrate some of its more important conceptual features and instill the physical meaning of many of the terms

Preface

used in the equations. These simplified applications increase interest in the subject matter and better prepare the student to understand the examples and solve problems.

Homework Problems. Apart from the Fundamental and Conceptual type problems mentioned previously, other types of problems contained in the book include the following:

• Free-Body Diagram Problems. Some sections of the book contain introductory problems that only require drawing the free-body diagram for the specific problems within a problem set. These assignments will impress upon the student the importance of mastering this skill as a requirement for a complete solution of any equilibrium problem. • General Analysis and Design Problems. The majority of problems in the book depict realistic situations encountered in engineering practice. Some of these problems come from actual products used in industry. It is hoped that this realism will both stimulate the student’s interest in engineering mechanics and provide a means for developing the skill to reduce any such problem from its physical description to a model or symbolic representation to which the principles of mechanics may be applied. An attempt has been made to arrange the problems in order of increasing difficulty except for the end of chapter review problems, which are presented in random order. • Computer Problems. An effort has been made to include some problems that may be solved using a numerical procedure executed on either a desktop computer or a programmable pocket calculator. The intent here is to broaden the student’s capacity for using other forms of mathematical analysis without sacrificing the time needed to focus on the application of the principles of mechanics. Problems of this type, which either can or must be solved using numerical procedures, are identified by a “square” symbol () preceding the problem number. The many homework problems in this edition, have been placed into two different categories. Problems that are simply indicated by a problem number have an answer and in some cases an additional numerical result given in the back of the book. An asterisk (*) before every fourth problem number indicates a problem without an answer.

Accuracy. As with the previous editions, apart from the author, the accuracy of the text and problem solutions has been thoroughly checked by four other parties: Scott Hendricks, Virginia Polytechnic Institute and State University; Karim Nohra, University of South Florida; Kurt Norlin, Bittner Development Group; and finally Kai Beng, a practicing engineer, who in addition to accuracy review provided suggestions for problem development.

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Animations. On the Companion Website are seven animations identified as fundamental engineering mechanics

concepts. The animations, flagged by a film icon, help students visualize the relation between mathematical explanation and real structure, breaking down complicated sequences and showing how free-body diagrams can be derived. These animations lend a graphic component in tutorials and lectures, assisting instructors in demonstrating the teaching of concepts with greater ease and clarity.

25 4

C h a p t e r 15

KinetiCs

of a

Website for the the Companion Masses Please refer to act of 2 Sliding animation: Imp

Each animation is flagged by a film icon.

pulse partiCle: im

and

momentum

tum

Linear Momen

rvation of icles 15.3 ConaseSy stem of Part for

on a system of impulses acting of the external m, namely, for d fie When the sum pli reduces to a sim zero, Eq. 15–6 (v )2 m i (vi)1 = m i i

particles is

(15–8)

momentum. tion of linear s as the conserva rticles remain is referred to a system of pa for m ntu This equation me = m ivi into r mo ea mv lin g G al tin tot itu the bst It states that riod t1 to t2 . Su g the time pe constant durin also write can we 8, 15– . Eq (15–9) (vG)1 = (v G)2

15

system of ss center for the system. ity v G of the ma ed to the that the veloc pulses are appli im al ern which indicates ext en particles not change if no mentum is often applied wh e-body particles does dy of the fre n of linear mo stu tio l rva efu nse car co a The in order to plication, uld be made eract. For ap sho int s or cle e rti pa llid co of pulses and entire system l or internal im d. diagram for the which create either externa m is conserve ntu me ces mo r for ea ion(s) lin s cancel identify the in what direct tem will alway ne sys mi the ter de for e y s thereb impulse pairs. If the tim r, the internal site collinear As stated earlie some of the equal but oppo rt, in sho r y cu oc ver y d is out, since the proximately tion is studie considered ap which the mo called neglected or period over le impulses are s may also be gib lse gli pu ne im l se na g the exter large and act forces causin which are very ces to zero. The for l , ua son me eq ari mo ntum to forces. By comp duce a significant change in in the top pho d in The hammer nonimpulsive not be neglecte d of time pro ulsive force rio can imp pe e, rt an urs sho co lies app for a very . They, of Du rin g thi s pulsive forces im led to the sta ke. cal of e are ng and rt tim sis. striki of ext rem ely sho omentum analy r due to an explosion or the weight of the the impulse–m ally occu may include the rm contact the ces no d for ere e ces sid lsiv for con Impulsive ereas nonimpu spring having sta ke  can be st another, wh htly deformed , and provided nonimpulsive one body again very small parted by a slig ven into soft y force that is dy, the force im an r, bo a tte of t ma the stake is dri t igh the we for tha s distinction or ulse of thi , g ess imp kin ffn the ma sti , en all ground a relatively sm lsive) forces. Wh to realize on the stake is important ground acting er larger (impu con sid ere d r the lsive forces, it compared to oth can   als o be strate, conside and nonimpu illu e st, if To lsiv tra . pu t con im to By n 2 t . betwee the time 1 oto. During nonimpulsive applies during shown in the ph d in a concrete ly as t on s ke ball thi rac t a the the stake is use tha ll with ak concrete, the racket on ng a tennis ba chipper to bre By , the force of effect of striki ulsive forces act m drastically. of interaction ntu e me tim rt mo then two imp sho change the very at its top due s the ball’s le effect on the ce it change on the stake: one gib sin er gli e oth ne lsiv a the pu ve r and ll ha is  im to the chippe ball’s weight wi to the bottom due comparison, the on its concrete. rigidity of the

Preface

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P r e fa c e

Video Solutions. An invaluable resource in and out of the classroom, these complete solution walkthroughs

of representative homework problems from each chapter offer fully worked solutions, self-paced instruction, and 24/7 accessibility. Lecturers and students can harness this resource to gain independent exposure to a wide range of examples applying formulas to actual structures.

icle: t r a P a s of c i t e n i rgy e K n E d n Work a

Video solutions are available for certain questions.

are lutions Video So for selected e r. availabl in this chapte ns questio

lve it to so d apply ergy an ent. and en m rk ce o la w p of d dis rinciple locity, an cy. p the p force, ve efficien develo e n To involve er and w at apply th o th p s e m volv rce and in fo proble at e iv th at lems. ms rv b le se ro b n p ro tic t of a co study p lve kine n To concep y to so ce the f energ introdu ation o rv se n To n m of co theore

ives

bjeCt

er O Chapt

a Force

of ncepts g the co solving cle usin r ti fo ar l p a fu tion of be use re we do this, yze mo uation will efo F ill anal ment. B g eq a force r, we w resultin ty, and displace Specifically, cement chapte e h is T th e. la ci In rc p rgy. lo e is fo n d ve a e a e, f d o s rc es n work a that involve fo ine the work undergo Fig. 14–1 cause s particle def in r, problem we must first only when the if the force F new position , , e e a le er cl th p to ti ev s, r ar w d am n ho r ex nap dr is positio work o f the force. Fo ude of n the s from will do e path r. The magnit angle betwee ty, on o ti th g ec n ir o d ti e al r in the h. If th move ar quan en dr = ticle to is a scal g the pat the par ent is th segment alon ork done by F placem the dis the differential –1, then the w of ig. 14 length F is u, F dr and tails of y ds cos u b dU = F defined

14.1

ork of The W

F

u

ds

dr r¿

r s

Fig. 14–1

Preface

W

Reduces lecturers’ time spent in repetitive explanation of concepts and applications.

Independent video replays of a lecturer’s explanation reinforces students’ understanding.

X

W

Flexible resource for students where they can learn at a comfortable pace without relying too much on their instuctors.

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Preface

Contents The book is divided into 11 chapters, in which the principles are first applied to simple, then to more complicated situations. The kinematics of a particle is discussed in Chapter 12, followed by a discussion of particle kinetics in Chapter 13 (Equation of Motion), Chapter 14 (Work and Energy), and Chapter 15 (Impulse and Momentum). The concepts of particle dynamics contained in these four chapters are then summarized in a “review” section, and the student is given the chance to identify and solve a variety of problems. A similar sequence of presentation is given for the planar motion of a rigid body: Chapter 16 (Planar Kinematics), Chapter 17 (Equations of Motion), Chapter 18 (Work and Energy), and Chapter 19 (Impulse and Momentum), followed by a summary and review set of problems for these chapters. If time permits, some of the material involving three-dimensional rigid-body motion may be included in the course. The kinematics and kinetics of this motion are discussed in Chapters 20 and 21, respectively. Chapter 22 (Vibrations) may be included if the student has the necessary mathematical background. Sections of the book that are considered to be beyond the scope of the basic dynamics course are indicated by a star () and may be omitted. Note that this material also provides a suitable reference for basic principles when it is discussed in more advanced courses. Finally, Appendix A provides a list of mathematical formulas needed to solve the problems in the book, Appendix B provides a brief review of vector analysis, and Appendix C reviews application of the chain rule.

Alternative Coverage. At the discretion of the instructor, it is possible to cover

Chapters 12 through 19 in the following order with no loss in continuity: Chapters 12 and 16 (Kinematics), Chapters 13 and 17 (Equations of Motion), Chapter 14 and 18 (Work and Energy), and Chapters 15 and 19 (Impulse and Momentum).

Acknowledgments The author has endeavored to write this book so that it will appeal to both the student and instructor. Through the years, many people have helped in its development, and I will always be grateful for their valued suggestions and comments. Specifically, I wish to thank all the individuals who have contributed their comments relative to preparing the fourteenth edition of this work, and in particular, R. Bankhead of Highline Community College, K. Cook-Chennault of Rutgers, the State University of New Jersey, E. Erisman, College of Lake County Illinois, M. Freeman of the University of Alabama, H. Lu of University of Texas at Dallas, J. Morgan of Texas A & M University, R. Neptune of the University of Texas, I. Orabi of the University of New Haven, T. Tan, University of Memphis, R. Viesca of Tufts University, and G. Young, Oklahoma State University. There are a few other people that I also feel deserve particular recognition. These include comments sent to me by J. Dix, H. Kuhlman, S. Larwood, D. Pollock, and H. Wenzel. A long-time friend and associate, Kai Beng Yap, was of great help to me in preparing and checking problem solutions. A special note of thanks also goes to

Preface

Kurt Norlin of Bittner Development Group in this regard. During the production process I am thankful for the assistance of Martha McMaster, my copy editor, and Rose Kernan, my production editor. Also, to my wife, Conny, who helped in the preparation of the manuscript for publication. Lastly, many thanks are extended to all my students and to members of the teaching profession who have freely taken the time to e-mail me their suggestions and comments. Since this list is too long to mention, it is hoped that those who have given help in this manner will accept this anonymous recognition. I would greatly appreciate hearing from you if at any time you have any comments, suggestions, or problems related to any matters regarding this edition. Russell Charles Hibbeler [email protected]

Global Edition The publishers would like to thank the following for their contribution to the Global Edition: Contributor Kai Beng Yap Kai is currently a registered professional engineer who works in Malaysia. He has BS and MS degrees in Civil Engineering from the University of Louisiana, Lafayette, Louisiana; and he has done further graduate work at Virginia Polytechnic Institute in Blacksberg, Virginia. His professional experience has involved teaching at the University of Louisiana, and doing engineering consulting work related to structural analysis and design and its associated infrastructure. Reviewers Derek Gransden, Structural Integrity & Composites / Faculty of Aerospace Engineering, Delft University of Technology Christopher Chin, National Centre for Maritime Engineering and Hydrodynamics, University of Tasmania Imad Abou-Hayt, Center of Bachelor of Engineering Studies, Technical University of Denmark Kris Henrioulle, Mechanical Engineering Technology, KU Leuven University

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your work...

your answer specific feedback

®

Resources for Instructors • MasteringEngineering. This online Tutorial Homework program allows you to integrate dynamic homework with automatic grading and adaptive tutoring. MasteringEngineering allows you to easily track the performance of your entire class on an assignment-by-assignment basis, or the detailed work of an individual student. • Instructor’s Solutions Manual. This supplement provides complete solutions supported by problem statements and problem figures. The fourteenth edition manual was revised to improve readability and was triple accuracy checked. The Instructor’s Solutions Manual is available on the Pearson Higher Education website: www.pearsonglobaleditions.com. • Instructor’s Resource. Visual resources to accompany the text are located on the Pearson Higher Education website: www.pearsonglobaleditions.com. If you are in need of a login and password for this site, please contact your local Pearson representative. Visual resources include all art from the text, available in PowerPoint and JPEG format. • Video Solutions. Developed by Professor Edward Berger, Purdue University, video solutions are located in the study area of MasteringEngineering and offer step-by-step solution walkthroughs of representative homework problems. Make efficient use of class time and office hours by showing students the complete and concise problemsolving approaches that they can access any time and view at their own pace. The videos are designed to be a flexible resource to be used however each instructor and student prefers. A valuable tutorial resource, the videos are also helpful for student self-evaluation as students can pause the videos to check their understanding and work alongside the video. Access the videos at www.masteringengineering.com

Resources for Students • MasteringEngineering. Tutorial homework problems emulate the instructor’s office-hour environment, guiding students through engineering concepts with self-paced individualized coaching. These in-depth tutorial homework problems are designed to coach students with feedback specific to their errors and optional hints that break problems down into simpler steps. • Dynamics Study Pack. This supplement contains chapter-by-chapter study materials and a Free-Body Diagram Workbook. • Video Solutions Complete, step-by-step solution walkthroughs of representative homework problems. Videos offer fully worked solutions that show every step of representative homework problems—this helps students make vital connections between concepts.

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Ordering Options The Dynamics Study Pack and MasteringEngineering resources are available as stand-alone items for student purchase and are also available packaged with the texts.

Custom Solutions Please contact your local Pearson sales representative for more details about custom options.

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CONTENTS 12

Kinematics of a Particle 2 Chapter Objectives 3 Introduction 3 Rectilinear Kinematics: Continuous Motion 5 12.3 Rectilinear Kinematics: Erratic Motion 20 12.4 General Curvilinear Motion 34 12.5 Curvilinear Motion: Rectangular Components 36 12.6 Motion of a Projectile 41 12.7 Curvilinear Motion: Normal and Tangential Components 56 12.8 Curvilinear Motion: Cylindrical Components 71 12.9 Absolute Dependent Motion Analysis of Two Particles 85 12.10 Relative-Motion of Two Particles Using Translating Axes 91

12.1 12.2

13

Kinetics of a Particle: Force and Acceleration 112 13.1 13.2 13.3 13.4 13.5 13.6 *13.7

Chapter Objectives 113 Newton’s Second Law of Motion 113 The Equation of Motion 116 Equation of Motion for a System of Particles 118 Equations of Motion: Rectangular Coordinates 120 Equations of Motion: Normal and Tangential Coordinates 138 Equations of Motion: Cylindrical Coordinates 152 Central-Force Motion and Space Mechanics 164

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contents

14

Kinetics of a Particle: Work and Energy 178 14.1 14.2 14.3 14.4 14.5 14.6

Chapter Objectives 179 The Work of a Force 179 Principle of Work and Energy 184 Principle of Work and Energy for a System of Particles 186 Power and Efficiency 204 Conservative Forces and Potential Energy 213 Conservation of Energy 217

15

Kinetics of a Particle: Impulse and Momentum 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 *15.9

236

Chapter Objectives 237 Principle of Linear Impulse and Momentum 237 Principle of Linear Impulse and Momentum for a System of Particles 240 Conservation of Linear Momentum for a System of Particles 254 Impact 266 Angular Momentum 280 Relation Between Moment of a Force and Angular Momentum 281 Principle of Angular Impulse and Momentum 284 Steady Flow of a Fluid Stream 295 Propulsion with Variable Mass 300

contents

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16

Planar Kinematics of a Rigid Body 318 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8

Chapter Objectives 319 Planar Rigid-Body Motion 319 Translation 321 Rotation about a Fixed Axis 322 Absolute Motion Analysis 338 Relative-Motion Analysis: Velocity 346 Instantaneous Center of Zero Velocity 360 Relative-Motion Analysis: Acceleration 373 Relative-Motion Analysis Using Rotating Axes 389

17

Planar Kinetics of a Rigid Body: Force and Acceleration 408 17.1 17.2 17.3 17.4 17.5

Chapter Objectives 409 Mass Moment of Inertia 409 Planar Kinetic Equations of Motion 423 Equations of Motion: Translation 426 Equations of Motion: Rotation about a Fixed Axis 441 Equations of Motion: General Plane Motion 456

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18

Planar Kinetics of a Rigid Body: Work and Energy 472 18.1 18.2 18.3 18.4 18.5

Chapter Objectives 473 Kinetic Energy 473 The Work of a Force 476 The Work of a Couple Moment 478 Principle of Work and Energy 480 Conservation of Energy 496

19

Planar Kinetics of a Rigid Body: Impulse and Momentum 516 19.1 19.2 19.3 *19.4

Chapter Objectives 517 Linear and Angular Momentum 517 Principle of Impulse and Momentum 523 Conservation of Momentum 540 Eccentric Impact 544

Preface

xxiii

20

Three-Dimensional Kinematics of a Rigid Body 560 20.1 *20.2 20.3 *20.4

Chapter Objectives 561 Rotation about a Fixed Point 561 The Time Derivative of a Vector Measured from Either a Fixed or Translating-Rotating System 564 General Motion 569 Relative-Motion Analysis Using Translating and Rotating Axes 578

21

Three-Dimensional Kinetics of a Rigid Body 590 *21.1 21.2 21.3 *21.4 *21.5 21.6

Chapter Objectives 591 Moments and Products of Inertia Angular Momentum 601 Kinetic Energy 604 Equations of Motion 612 Gyroscopic Motion 626 Torque-Free Motion 632

591

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22 Vibrations 642 *22.1 *22.2 *22.3 *22.4 *22.5 *22.6

Chapter Objectives 643 Undamped Free Vibration 643 Energy Methods 657 Undamped Forced Vibration 663 Viscous Damped Free Vibration 667 Viscous Damped Forced Vibration 670 Electrical Circuit Analogs 673

Appendix A. B. C.

Mathematical Expressions Vector Analysis 684 The Chain Rule 689

Fundamental Problems Partial Solutions and Answers 692 Preliminary Problems Dynamics Solutions 713 Review Problem Solutions 723 Answers to Selected Problems 733 Index

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CREDITS Chapter opening images are credited as follows: Chapter 12, Lars Johansson/Fotolia Chapter 13, Migel/Shutterstock Chapter 14, Oliver Furrer/Ocean/Corbis Chapter 15, David J. Green/Alamy Chapter 16, TFoxFoto/Shutterstock Chapter 17, Surasaki/Fotolia Chapter 18, Arinahabich/Fotolia Chapter 19, Hellen Sergeyeva/Fotolia Chapter 20, Philippe Psaila/Science Source Chapter 21, Derek Watt/Alamy Chapter 22, Daseaford/Fotolia

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ENGINEERING MECHANICS

DYNAMICS FourTEENTH EDITIoN IN sI uNITs

Chapter 12

(© Lars Johansson/Fotolia) Although each of these boats is rather large, from a distance their motion can be analyzed as if each were a particle.

Kinematics of a Particle Chapter ObjeCtives n

To introduce the concepts of position, displacement, velocity, and acceleration.

n

To study particle motion along a straight line and represent this motion graphically.

n

To investigate particle motion along a curved path using different coordinate systems.

n

To present an analysis of dependent motion of two particles.

n

To examine the principles of relative motion of two particles using translating axes.

12.1

Introduction

Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies subjected to the action of forces. Engineering mechanics is divided into two areas of study, namely, statics and dynamics. Statics is concerned with the equilibrium of a body that is either at rest or moves with constant velocity. Here we will consider dynamics, which deals with the accelerated motion of a body. The subject of dynamics will be presented in two parts: kinematics, which treats only the geometric aspects of the motion, and kinetics, which is the analysis of the forces causing the motion. To develop these principles, the dynamics of a particle will be discussed first, followed by topics in rigid-body dynamics in two and then three dimensions.

Video Solutions are available for selected questions in this chapter.

4

12

Chapter 12

K i n e m at i C s

of a

partiCle

Historically, the principles of dynamics developed when it was possible to make an accurate measurement of time. Galileo Galilei (1564–1642) was one of the first major contributors to this field. His work consisted of experiments using pendulums and falling bodies. The most significant contributions in dynamics, however, were made by Isaac Newton (1642–1727), who is noted for his formulation of the three fundamental laws of motion and the law of universal gravitational attraction. Shortly after these laws were postulated, important techniques for their application were developed by Euler, D’Alembert, Lagrange, and others. There are many problems in engineering whose solutions require application of the principles of dynamics. Typically the structural design of any vehicle, such as an automobile or airplane, requires consideration of the motion to which it is subjected. This is also true for many mechanical devices, such as motors, pumps, movable tools, industrial manipulators, and machinery. Furthermore, predictions of the motions of artificial satellites, projectiles, and spacecraft are based on the theory of dynamics. With further advances in technology, there will be an even greater need for knowing how to apply the principles of this subject.

Problem Solving.

Dynamics is considered to be more involved than statics since both the forces applied to a body and its motion must be taken into account. Also, many applications require using calculus, rather than just algebra and trigonometry. In any case, the most effective way of learning the principles of dynamics is to solve problems. To be successful at this, it is necessary to present the work in a logical and orderly manner as suggested by the following sequence of steps: 1. Read the problem carefully and try to correlate the actual physical situation with the theory you have studied. 2. Draw any necessary diagrams and tabulate the problem data. 3. Establish a coordinate system and apply the relevant principles, generally in mathematical form. 4. Solve the necessary equations algebraically as far as practical; then, use a consistent set of units and complete the solution numerically. Report the answer with no more significant figures than the accuracy of the given data. 5. Study the answer using technical judgment and common sense to determine whether or not it seems reasonable. 6. Once the solution has been completed, review the problem. Try to think of other ways of obtaining the same solution. In applying this general procedure, do the work as neatly as possible. Being neat generally stimulates clear and orderly thinking, and vice versa.

12.2

12.2

5

reCtilinear KinematiCs: Continuous motion

Rectilinear Kinematics: Continuous Motion

12

We will begin our study of dynamics by discussing the kinematics of a particle that moves along a rectilinear or straight-line path. Recall that a particle has a mass but negligible size and shape. Therefore we must limit application to those objects that have dimensions that are of no consequence in the analysis of the motion. In most problems, we will be interested in bodies of finite size, such as rockets, projectiles, or vehicles. Each of these objects can be considered as a particle, as long as the motion is characterized by the motion of its mass center and any rotation of the body is neglected.

Rectilinear Kinematics. The kinematics of a particle is characterized by specifying, at any given instant, the particle’s position, velocity, and acceleration. Position.

The straight-line path of a particle will be defined using a single coordinate axis s, Fig. 12–1a. The origin O on the path is a fixed point, and from this point the position coordinate s is used to specify the location of the particle at any given instant. The magnitude of s is the distance from O to the particle, usually measured in meters (m), and the sense of direction is defined by the algebraic sign on s. Although the choice is arbitrary, in this case s is positive since the coordinate axis is positive to the right of the origin. Likewise, it is negative if the particle is located to the left of O. Realize that position is a vector quantity since it has both magnitude and direction. Here, however, it is being represented by the algebraic scalar s, rather than in boldface s, since the direction always remains along the coordinate axis.

s

O s Position (a)

Displacement.

The displacement of the particle is defined as the change in its position. For example, if the particle moves from one point to another, Fig. 12–1b, the displacement is

s

O s

s s¿

s = s - s

Displacement (b)

In this case s is positive since the particle’s final position is to the right of its initial position, i.e., s 7 s. Likewise, if the final position were to the left of its initial position, s would be negative. The displacement of a particle is also a vector quantity, and it should be distinguished from the distance the particle travels. Specifically, the distance traveled is a positive scalar that represents the total length of path over which the particle travels.

Fig. 12–1

6

Chapter 12

K i n e m at i C s

of a

partiCle

Velocity. If the particle moves through a displacement s during the

12

time interval t, the average velocity of the particle during this time interval is s t

vavg =

If we take smaller and smaller values of t, the magnitude of s becomes smaller and smaller. Consequently, the instantaneous velocity is a vector defined as v = lim (s> t), or S t

0

+ ) (S

v s

O s Velocity (c)

v =

ds dt

(12–1)

Since t or dt is always positive, the sign used to define the sense of the velocity is the same as that of s or ds. For example, if the particle is moving to the right, Fig. 12–1c, the velocity is positive; whereas if it is moving to the left, the velocity is negative. (This is emphasized here by the arrow written at the left of Eq. 12–1.) The magnitude of the velocity is known as the speed, and it is generally expressed in units of m>s. Occasionally, the term “average speed” is used. The average speed is always a positive scalar and is defined as the total distance traveled by a particle, sT , divided by the elapsed time t; i.e.,

(vsp)avg =

sT t

For example, the particle in Fig. 12–1d travels along the path of length sT in time t, so its average speed is (vsp)avg = sT > t, but its average velocity is vavg = - s> t. s P¿

P

O sT Average velocity and Average speed (d)

Fig. 12–1 (cont.)

s

12.2

7

reCtilinear KinematiCs: Continuous motion

Acceleration. Provided the velocity of the particle is known at two  points, the average acceleration of the particle during the time interval t is defined as v t

aavg =

Here v represents the difference in the velocity during the time interval t, i.e., v = v - v, Fig. 12–1e. The instantaneous acceleration at time t is a vector that is found by taking smaller and smaller values of t and corresponding smaller and smaller values of v, so that a = lim (v> t), or S t

12

0

a s

O v

+ ) (S

dv a = dt

v¿

Acceleration

(12–2)

(e)

Substituting Eq. 12–1 into this result, we can also write + ) (S

a =

d2s dt2

Both the average and instantaneous acceleration can be either positive or negative. In particular, when the particle is slowing down, or its speed is decreasing, the particle is said to be decelerating. In this case, v in Fig. 12–1f is less than v, and so v = v - v will be negative. Consequently, a will also be negative, and therefore it will act to the left, in the opposite sense to v. Also, notice that if the particle is originally at rest, then it can have an acceleration if a moment later it has a velocity v; and, if the velocity is constant, then the acceleration is zero since v = v - v = 0. Units commonly used to express the magnitude of acceleration are m>s2. Finally, an important differential relation involving the displacement, velocity, and acceleration along the path may be obtained by eliminating the time differential dt between Eqs. 12–1 and 12–2. We have dt =

ds dv = v a

or + ) (S

a ds = v dv

(12–3)

Although we have now produced three important kinematic equations, realize that the above equation is not independent of Eqs. 12–1 and 12–2.

a P

P¿

O v Deceleration (f)

Fig. 12–1 (cont.)

v¿

s

8

Chapter 12

K i n e m at i C s

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Constant Acceleration, a = ac . When the acceleration is constant, each of the three kinematic equations ac = dv>dt, v = ds>dt, and ac ds = v dv can be integrated to obtain formulas that relate ac , v, s, and t.

12

Velocity as a Function of Time.

Integrate ac = dv>dt, assuming

that initially v = v0 when t = 0. v

When the ball is released, it has zero velocity but an acceleration of 9.81 m>s2.

Lv0

L0

dv =

t

ac dt

v = v0 + ac t

+ ) (S

Constant Acceleration

(12–4)

Position as a Function of Time. Integrate v = ds>dt = v0 + act, assuming that initially s = s0 when t = 0. s

Ls0 + ) (S

ds =

L0

t

(v0 + act) dt

s = s0 + v0t + 12 ac t2 Constant Acceleration

(12–5)

Velocity as a Function of Position. Either solve for t in

Eq. 12–4 and substitute into Eq. 12–5, or integrate v dv = ac ds, assuming that initially v = v0 at s = s0 . v

Lv0 + ) (S

v dv =

s

Ls0

ac ds

v2 = v20 + 2ac(s - s0) Constant Acceleration

(12–6)

The algebraic signs of s0 , v0 , and ac , used in the above three equations, are determined from the positive direction of the s axis as indicated by the arrow written at the left of each equation. Remember that these equations are useful only when the acceleration is constant and when t = 0, s = s0 , v = v0 . A typical example of constant accelerated motion occurs when a body falls freely toward the earth. If air resistance is neglected and the distance of fall is short, then the downward acceleration of the body when it is close to the earth is constant and approximately 9.81 m>s2. The proof of this is given in Example 13.2.

12.2

reCtilinear KinematiCs: Continuous motion

important points • • • • • •

12

Dynamics is concerned with bodies that have accelerated motion. Kinematics is a study of the geometry of the motion. Kinetics is a study of the forces that cause the motion. Rectilinear kinematics refers to straight-line motion. Speed refers to the magnitude of velocity. Average speed is the total distance traveled divided by the total time. This is different from the average velocity, which is the displacement divided by the time.

• A particle that is slowing down is decelerating. • A particle can have an acceleration and yet have zero velocity. • The relationship a ds = v dv is derived from a = dv>dt and v = ds>dt, by eliminating dt.

During the time this rocket undergoes rectilinear motion, its altitude as a function of time can be measured and expressed as s = s(t). Its velocity can then be found using v = ds>dt, and its acceleration can  be determined from a = dv>dt. (© NASA)

procedure for analysis Coordinate System. • Establish a position coordinate s along the path and specify its fixed origin and positive direction.

• Since motion is along a straight line, the vector quantities position, velocity, and acceleration can be

represented as algebraic scalars. For analytical work the sense of s, v, and a is then defined by their algebraic signs.

• The positive sense for each of these scalars can be indicated by an arrow shown alongside each kinematic equation as it is applied. Kinematic Equations.

• If a relation is known between any two of the four variables a, v, s, and t, then a third variable can be

obtained by using one of the kinematic equations, a = dv>dt, v = ds>dt or a ds = v dv, since each equation relates all three variables.*

• Whenever integration is performed, it is important that the position and velocity be known at a given

instant in order to evaluate either the constant of integration if an indefinite integral is used, or the limits of integration if a definite integral is used.

• Remember that Eqs. 12–4 through 12–6 have only limited use. These equations apply only when the acceleration is constant and the initial conditions are s = s0 and v = v0 when t = 0.

*Some standard differentiation and integration formulas are given in Appendix A.

9

10

12

Chapter 12

ExamplE

K i n e m at i C s

of a

partiCle

12.1 The car on the left in the photo and in Fig. 12–2 moves in a straight line such that for a short time its velocity is defined by v = (0.6t2 + t) m>s, where t is in seconds. Determine its position and acceleration when t = 3 s. When t = 0, s = 0. a, v

s

O

Fig. 12–2

Solution Coordinate system. The position coordinate extends from the fixed origin O to the car, positive to the right. position. Since v = f(t), the car’s position can be determined from v = ds>dt, since this equation relates v, s, and t. Noting that s = 0 when t = 0, we have* + ) (S

v = L0

s

ds = s`

s 0

ds = (0.6t2 + t) dt L0

t

(0.6t2 + t)dt

= 0.2t3 + 0.5t2 ` 3

t 0

2

s = (0.2t + 0.5t ) m

When t = 3 s, s = 0.2(3)3 + 0.5(3)2 = 9.90 m

Ans.

acceleration. Since v = f(t), the acceleration is determined from a = dv>dt, since this equation relates a, v, and t. + ) (S

dv d = (0.6t2 + t) dt dt = (1.2t + 1) m>s2

a =

When t = 3 s, a = 1.2(3) + 1 = 4.60 m>s2 S

Ans.

NOTE: The formulas for constant acceleration cannot be used to solve

this problem, because the acceleration is a function of time. *The same result can be obtained by evaluating a constant of integration C rather than using definite limits on the integral. For example, integrating ds = (0.6t2 + t)dt yields s = 0.2t3 + 0.5t2 + C. Using the condition that at t = 0, s = 0, then C = 0.

12.2

ExamplE

reCtilinear KinematiCs: Continuous motion

12.2

12

A small projectile is fired vertically downward into a fluid medium with an initial velocity of 60 m>s. Due to the drag resistance of the fluid the projectile experiences a deceleration of a = ( -0.4v3) m>s2, where v is in m>s. Determine the projectile’s velocity and position 4 s after it is fired. Solution Coordinate system. Since the motion is downward, the position coordinate is positive downward, with origin located at O, Fig. 12–3. velocity. Here a = f(v) and so we must determine the velocity as a function of time using a = dv>dt, since this equation relates v, a, and t. (Why not use v = v0 + act?) Separating the variables and integrating, with v0 = 60 m>s when t = 0, yields dv = -0.4v3 dt v t dv = dt 3 L60 m>s -0.4v L0

(+ T )

a =

position. Knowing v = f(t), we can obtain the projectile’s position from v = ds>dt, since this equation relates s, v, and t. Using the initial condition s = 0, when t = 0, we have v = L0

s

-1>2 1 + 0.8t d dt 2 L0 (60) t

c

1>2 t 2 1 c + 0.8t d ` 0.8 (60)2 0 1>2 1 1 1 s = ec + 0.8t d fm 0.4 (60)2 60

s =

When t = 4 s,

-1>2 ds 1 = c + 0.8t d dt (60)2

ds =

s = 4.43 m

O s

Fig. 12–3

1 1 1 v b 2` = t - 0 a -0.4 -2 v 60 1 1 1 c 2 d = t 0.8 v (60)2 -1>2 1 v = ec + 0.8t d f m>s (60)2 Here the positive root is taken, since the projectile will continue to move downward. When t = 4 s, v = 0.559 m>s T Ans.

(+ T )

11

Ans.

12

12

Chapter 12

K i n e m at i C s

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12.3

ExamplE

During a test a rocket travels upward at 75 m>s, and when it is 40 m from the ground its engine fails. Determine the maximum height sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m>s2 due to gravity. Neglect the effect of air resistance. Solution Coordinate system. The origin O for the position coordinate s is taken at ground level with positive upward, Fig. 12–4. Maximum height. Since the rocket is traveling upward, vA = +75 m>s when t = 0. At the maximum height s = sB the velocity vB = 0. For the entire motion, the acceleration is ac = -9.81 m>s2 (negative since it acts in the opposite sense to positive velocity or positive displacement). Since ac is constant the rocket’s position may be related to its velocity at the two points A and B on the path by using Eq. 12–6, namely,

vB  0 B

(+ c )

v2B = v2A + 2ac(sB - sA) 0 = (75 m>s)2 + 2( -9.81 m>s2)(sB - 40 m) sB = 327 m

sB

Ans.

velocity. To obtain the velocity of the rocket just before it hits the ground, we can apply Eq. 12–6 between points B and C, Fig. 12–4.

vA  75 m/s

(+ c )

A

v2C = v2B + 2ac(sC - sB) = 0 + 2(-9.81 m>s2)(0 - 327 m) vC = -80.1 m>s = 80.1 m>s T

sA  40 m C

Fig. 12–4

s

O

Ans.

The negative root was chosen since the rocket is moving downward. Similarly, Eq. 12–6 may also be applied between points A and C, i.e., (+ c )

v2C = v2A + 2ac(sC - sA) = (75 m>s)2 + 2(-9.81 m>s2)(0 - 40 m) vC = -80.1 m>s = 80.1 m>s T

Ans.

NOTE: It should be realized that the rocket is subjected to a deceleration

from A to B of 9.81 m>s2, and then from B to C it is accelerated at this rate. Furthermore, even though the rocket momentarily comes to rest at B (vB = 0) the acceleration at B is still 9.81 m>s2 downward!

12.2

ExamplE

13

reCtilinear KinematiCs: Continuous motion

12.4

12

A metallic particle is subjected to the influence of a magnetic field as it travels downward through a fluid that extends from plate A to plate B, Fig. 12–5. If the particle is released from rest at the midpoint C, s = 100 mm, and the acceleration is a = (4s) m>s2, where s is in meters, determine the velocity of the particle when it reaches plate B, s = 200 mm, and the time it takes to travel from C to B. Solution Coordinate system. As shown in Fig. 12–5, s is positive downward, measured from plate A. velocity. Since a = f(s), the velocity as a function of position can be obtained by using v dv = a ds. Realizing that v = 0 at s = 0.1 m, we have (+ T )

v

100 mm

v dv = a ds

s

s

L0 L0.1 m v 1 2 4 s v ` = s2 ` 2 2 0.1 m 0 v dv =

2

C

4s ds

1>2

v = 2(s - 0.01)

m>s

vB = 0.346 m>s = 346 mm>s T

(1) Ans.

The positive root is chosen since the particle is traveling downward, i.e., in the +s direction. time. The time for the particle to travel from C to B can be obtained using v = ds>dt and Eq. 1, where s = 0.1 m when t = 0. From Appendix A, ds = v dt = 2(s2 - 0.01)1>2dt t ds = 2 dt L0.1 (s2 - 0.01)1>2 L0 s

At s = 0.2 m,

ln 1 2s2 - 0.01 + s 2 `

s 0.1

= 2t `

t 0

ln 1 2s2 - 0.01 + s 2 + 2.303 = 2t

ln 1 2(0.2)2 - 0.01 + 0.2 2 + 2.303

= 0.658 s Ans. 2 NOTE: The formulas for constant acceleration cannot be used here because the acceleration changes with position, i.e., a = 4s. t =

200 mm

B

At s = 200 mm = 0.2 m,

(+ T )

A

Fig. 12–5

14

12

Chapter 12

K i n e m at i C s

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partiCle

12.5

ExamplE

A particle moves along a horizontal path with a velocity of v = (3t2 - 6t) m>s, where t is the time in seconds. If it is initially located at the origin O, determine the distance traveled in 3.5 s, and the particle’s average velocity and average speed during the time interval.

s  4.0 m

s  6.125 m O

t2s

t0s

t  3.5 s

(a)

Solution Coordinate system. Here positive motion is to the right, measured from the origin O, Fig. 12–6a. Distance traveled. Since v = f(t), the position as a function of time may be found by integrating v = ds>dt with t = 0, s = 0. + ) (S ds = v dt = (3t2 - 6t) dt L0

v (m/s) v  3t2  6t (0, 0)

(2 s, 0)

(1 s, 3 m/s) (b)

t (s)

s

t

(3t2 - 6t) dt L0 s = (t3 - 3t2) m

ds =

(1)

In order to determine the distance traveled in 3.5 s, it is necessary to investigate the path of motion. If we consider a graph of the velocity function, Fig. 12–6b, then it reveals that for 0 6 t 6 2 s the velocity is negative, which means the particle is traveling to the left, and for t 7 2 s the velocity is positive, and hence the particle is traveling to the right. Also, note that v = 0 at t = 2 s. The particle’s position when t = 0, t = 2 s, and t = 3.5 s can be determined from Eq. 1. This yields s t = 0 = 0 s t = 2 s = -4.0 m s t = 3.5 s = 6.125 m

Fig. 12–6

The path is shown in Fig. 12–6a. Hence, the distance traveled in 3.5 s is sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m velocity.

Ans.

The displacement from t = 0 to t = 3.5 s is s = s t = 3.5 s - s t = 0 = 6.125 m - 0 = 6.125 m

and so the average velocity is s 6.125 m = = 1.75 m>s S Ans. t 3.5 s - 0 The average speed is defined in terms of the distance traveled sT . This positive scalar is vavg =

(vsp)avg =

sT 14.125 m = = 4.04 m>s t 3.5 s - 0

Ans.

NOTE: In this problem, the acceleration is a = dv>dt = (6t - 6) m>s2,

which is not constant.

12.2

15

reCtilinear KinematiCs: Continuous motion

It is highly suggested that you test yourself on the solutions to these examples, by covering them over and then trying to think about which equations of kinematics must be used and how they are applied in order to determine the unknowns. Then before solving any of the problems, try and solve some of the Preliminary and Fundamental Problems which follow. The solutions and answers to all these problems are given in the back of the book. Doing this throughout the book will help immensely in understanding how to apply the theory, and thereby develop your problem-solving skills.

12

preliMinary prObleM P12–1. a) If s = (2t3) m, where t is in seconds, determine v when t = 2 s.

g) If a = 4 m > s2, determine s when t = 3 s if v = 2 m > s and s = 2 m when t = 0.

b) If v = (5s) m > s, where s is in meters, determine a at s = 1 m.

h) If a = (8t2) m > s2, determine v when t = 1 s if v = 0 at t = 0.

c) If v = (4t + 5) m > s, where t is in seconds, determine a when t = 2 s.

i) If s = (3t2 + 2) m, determine v when t = 2 s.

d) If a = 2 m > s2, determine v when t = 2 s if v = 0 when t = 0.

j) When t = 0 the particle is at A. In four seconds it travels to B, then in another six seconds it travels to C. Determine the average velocity and the average speed. The origin of the coordinate is at O.

e) If a = 2 m > s2, determine v at s = 4 m if v = 3 m > s at s = 0.

7m

1m A

O

B 14 m

m > s2,

f) If a = (s) where s is in meters, determine v when s = 5 m if v = 0 at s = 4 m.

Prob. P12–1

C

s

16

12

Chapter 12

K i n e m at i C s

of a

partiCle

FunDaMental prObleMs F12–1. Initially, the car travels along a straight road with a speed of 35 m>s. If the brakes are applied and the speed of the car is reduced to 10 m>s in 15 s, determine the constant deceleration of the car.

F12–5. The position of the particle is given by s = (2t2 - 8t + 6) m, where t is in seconds. Determine the time when the velocity of the particle is zero, and the total distance traveled by the particle when t = 3 s.

s

Prob. F12–1

F12–2. A ball is thrown vertically upward with a speed of 15 m>s. Determine the time of flight when it returns to its original position.

Prob. F12–5

F12–6. A particle travels along a straight line with an acceleration of a = (10 - 0.2s) m>s2, where s is measured in meters. Determine the velocity of the particle when s = 10 m if v = 5 m>s at s = 0.

s

s s

Prob. F12–6 Prob. F12–2

F12–3. A particle travels along a straight line with a velocity of v = (4t - 3t2) m>s, where t is in seconds. Determine the position of the particle when t = 4 s. s = 0 when t = 0.

F12–7. A particle moves along a straight line such that its acceleration is a = (4t2 - 2) m>s2, where t is in seconds. When t = 0, the particle is located 2 m to the left of the origin, and when t = 2 s, it is 20 m to the left of the origin. Determine the position of the particle when t = 4 s.

F12–4. A particle travels along a straight line with a speed v = (0.5t3 - 8t) m>s, where t is in seconds. Determine the acceleration of the particle when t = 2 s.

F12–8. A particle travels along a straight line with a velocity of v = (20 - 0.05s2) m>s, where s is in meters. Determine the acceleration of the particle at s = 15 m.

12.2

17

reCtilinear KinematiCs: Continuous motion

prObleMs 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m>s2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s? 12–2. The acceleration of a particle as it moves along a straight line is given by a = (4t3 - 1) m>s2, where t is in seconds. If s = 2 m and v = 5 m>s when t = 0, determine the particle’s velocity and position when t = 5 s. Also, determine the total distance the particle travels during this time period.

12 12–9. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB, respectively. If B maintains its constant speed, while A begins to decelerate at aA, determine the distance d between the cars at the instant A stops.

A

B

d

12–3. The velocity of a particle traveling in a straight line is given by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-s time interval, and what is its average speed?

Prob. 12–9

*12–4. A particle is moving along a straight line such that its position is defined by s = (10t2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s.

12–10. A particle travels along a straight-line path such that in 4 s it moves from an initial position sA = -8 m to a position sB = +3 m. Then in another 5 s it moves from sB to sC = -6 m. Determine the particle’s average velocity and average speed during the 9-s time interval.

12–5. A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. 12–6. A stone A is dropped from rest down a well, and in 1  s another stone B is dropped from rest. Determine the distance between the stones another second later. 12–7. A bus starts from rest with a constant acceleration of 1 m>s2. Determine the time required for it to attain a speed of 25 m>s and the distance traveled. *12–8. A particle travels along a straight line with a velocity v = (12 - 3t2) m>s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period.

12–11. Traveling with an initial speed of 70 km>h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km>h? Also, through what distance does the car travel during this time?

*12–12. A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m. Use a numerical method to evaluate the integral.

12–13. The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.

18

Chapter 12

K i n e m at i C s

of a

partiCle

12–14. A train starts from rest at station A and accelerates 12 at 0.5  m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations.

*12–20. The acceleration of a rocket traveling upward is given by a = (6 + 0.02s) m>s2, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0.

12–15. A particle is moving along a straight line such that its velocity is defined as v = (- 4s2) m>s, where s is in meters. If s = 2 m when t = 0, determine the velocity and acceleration as functions of time. *12–16. Determine the time required for a car to travel 1  km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m>s2 and decelerate at 2 m>s2. 12–17. A particle is moving with a velocity of v0 when s = 0 and t = 0. If it is subjected to a deceleration of a = -kv3, where k is a constant, determine its velocity and position as functions of time. s

12–18. A particle is moving along a straight line with an initial velocity of 6 m>s when it is subjected to a deceleration of a = (-1.5v1> 2) m>s2, where v is in m>s. Determine how far it travels before it stops. How much time does this take?

Prob. 12–20

12–19. The acceleration of a rocket traveling upward is given by a = (6 + 0.02s) m>s2, where s is in meters. Determine the rocket’s velocity when s = 2 km and the time needed to reach this attitude. Initially, v = 0 and s = 0 when t = 0.

12–21. When a train is traveling along a straight track at 2 m>s, it begins to accelerate at a = (60v-4) m>s2, where v is in m>s. Determine its velocity v and the position 3 s after the acceleration.

s

Prob. 12–19

v

s

Prob. 12–21

12.2 12–22. The acceleration of a particle along a straight line is defined by a = (2t - 9) m>s2, where t is in seconds. At t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity.

12–23. If the effects of atmospheric resistance are accounted for, a freely falling body has an acceleration defined by the equation a = 9.81[1 - v 2(10 -4)] m>s2, where v is in m>s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as t S  ).

*12–24. A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m>s. If the bag is released with the same upward velocity of 6 m>s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant.

12–25. A particle is moving along a straight line such that its acceleration is defined as a = (-2v) m>s2, where v is in meters per second. If v = 20 m>s when s = 0 and t = 0, determine the particle’s position, velocity, and acceleration as functions of time.

12–26. The acceleration of a particle traveling along a 1 straight line is a = s1>2 m>s2, where s is in meters. If v = 0, 4 s = 1 m when t = 0, determine the particle’s velocity at s = 2 m.

12–27. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf . If this variation of the acceleration can be expressed as a = (g>v2f) (v2f - v2), determine the time needed for the velocity to become v = vf >2. Initially the particle falls from rest. *12–28. A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = (- 6t) m>s2, where t is in seconds, determine the distance traveled before it stops.

reCtilinear KinematiCs: Continuous motion

19

12–29. A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At 12 the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m>s. Determine the height from the ground and the time at which they pass.

12–30. A boy throws a ball straight up from the top of a 12-m high tower. If the ball falls past him 0.75 s later, determine the velocity at which it was thrown, the velocity of the ball when it strikes the ground, and the time of flight.

12–31. The velocity of a particle traveling along a straight line is v = v0 - ks, where k is constant. If s = 0 when t = 0, determine the position and acceleration of the particle as a function of time.

*12–32. Ball A is thrown vertically upwards with a velocity of v0. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t < 2v0 >g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

12–33. As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = -g0[R 2 >(R + y)2], where g0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g0 = 9.81 m>s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth. Hint: This requires that v = 0 as y S  .

12–34. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–40), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y 0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y 0 = 500 km? Use the numerical data in Prob. 12–33.

20

Chapter 12

K i n e m at i C s

s

12

ds ds v0  dt t  0 v2  dt t 2 ds ds v1  dt t v3  dt t 1 3 s2

s1 O

t1

s3

t2

t

t3

(a)

v

of a

partiCle

12.3

Rectilinear Kinematics: Erratic Motion

When a particle has erratic or changing motion then its position, velocity, and acceleration cannot be described by a single continuous mathematical function along the entire path. Instead, a series of functions will be required to specify the motion at different intervals. For this reason, it is convenient to represent the motion as a graph. If a graph of the motion that relates any two of the variables s,v, a, t can be drawn, then this graph can be used to construct subsequent graphs relating two other variables since the variables are related by the differential relationships v = ds>dt, a = dv>dt, or a ds = v dv. Several situations occur frequently.

The s–t, v–t, and a–t Graphs. To construct the v9t graph given the s–t graph, Fig. 12–7a, the equation v = ds>dt should be used, since it relates the variables s and t to v. This equation states that v1

v0

v2 O

t1

t2

t3 v3

t

(b)

ds = v dt slope of = velocity s9t graph

Fig. 12–7

v a  dv a0  dv dt t  0 2 dt t2 dv a3  dt t a1  dv 3 dt t1

dv = a dt

v3 v2 v1

v0 O

t1

t2

t

t3

(a)

a

a1 O

a0  0

t1

a2 t2 (b)

Fig. 12–8

a3 t3

For example, by measuring the slope on the s–t graph when t = t1, the velocity is v1, which is plotted in Fig. 12–7b. The v9t graph can be constructed by plotting this and other values at each instant. The a–t graph can be constructed from the v9t graph in a similar manner, Fig. 12–8, since

t

slope of = acceleration v9t graph Examples of various measurements are shown in Fig. 12–8a and plotted in Fig. 12–8b. If the s–t curve for each interval of motion can be expressed by a mathematical function s = s(t), then the equation of the v9t graph for the same interval can be obtained by differentiating this function with respect to time since v = ds/dt. Likewise, the equation of the a–t graph for the same interval can be determined by differentiating v = v(t) since a = dv>dt. Since differentiation reduces a polynomial of degree n to that of degree n – 1, then if the s–t graph is parabolic (a second-degree curve), the v9t graph will be a sloping line (a first-degree curve), and the a–t graph will be a constant or a horizontal line (a zero-degree curve).

12.3

If the a–t graph is given, Fig. 12–9a, the v9t graph may be constructed using a = dv>dt, written as

a

12

a0

t1

v   a dt 0

L change in area under = velocity a9t graph v =

21

reCtilinear KinematiCs: erratiC motion

Hence, to construct the v9t graph, we begin with the particle’s initial velocity v0 and then add to this small increments of area (v) determined from the a–t graph. In this manner successive points, v1 = v0 + v, etc., for the v9t graph are determined, Fig. 12–9b. Notice that an algebraic addition of the area increments of the a–t graph is necessary, since areas lying above the t axis correspond to an increase in v (“positive” area), whereas those lying below the axis indicate a decrease in v (“negative” area). Similarly, if the v9t graph is given, Fig. 12–10a, it is possible to determine the s–t graph using v = ds>dt, written as

t

t1

a dt

(a) v

v

v1 v0

t

t1 (b)

Fig. 12–9

v

v dt L area under displacement = v9t graph s =

v0 t1

s   v dt 0

t

t1 (a)

In the same manner as stated above, we begin with the particle’s initial position s0 and add (algebraically) to this small area increments s determined from the v9t graph, Fig. 12–10b. If segments of the a–t graph can be described by a series of equations, then each of these equations can be integrated to yield equations describing the corresponding segments of the v9t graph. In a similar manner, the s–t graph can be obtained by integrating the equations which describe the segments of the v9t graph. As a result, if the a–t graph is linear (a first-degree curve), integration will yield a v9t graph that is parabolic (a second-degree curve) and an s–t graph that is cubic (third-degree curve).

s

s1

s

s0 t

t1 (b)

Fig. 12–10

22

Chapter 12

K i n e m at i C s

of a

partiCle

The v–s and a–s Graphs. If the a–s graph can be constructed,

a

12

then points on the v9s graph can be determined by using v dv = a ds. Integrating this equation between the limits v = v0 at s = s0 and v = v1 at s = s1 , we have,

s1

a0

0 a ds  —12 (v12  v02) s

s1 (a) v

s1

1 2 2 (v1

-

v20)

a ds Ls0 area under a9s graph

=

v1 v0

s

s1

Therefore, if the red area in Fig. 12–11a is determined, and the initial velocity v0 at s0 = 0 is known, then v1 = 1 2 10s1a ds + v20 2 1>2, Fig. 12–11b. Successive points on the v–s graph can be constructed in this manner. If the v–s graph is known, the acceleration a at any position s can be determined using a ds = v dv, written as

(b)

Fig. 12–11 v dv ds

v0

v

s

s (a) a

dv b ds velocity times acceleration = slope of v9s graph a = va

a0 a  v(dv/ds) s

s (b)

Fig. 12–12

Thus, at any point (s, v) in Fig. 12–12a, the slope dv>ds of the v–s graph is measured. Then with v and dv>ds known, the value of a can be calculated, Fig. 12–12b. The v–s graph can also be constructed from the a–s graph, or vice versa, by approximating the known graph in various intervals with mathematical functions, v = f(s) or a = g(s), and then using a ds = v dv to obtain the other graph.

12.3

23

reCtilinear KinematiCs: erratiC motion

12.6

ExamplE

12

A bicycle moves along a straight road such that its position is described by the graph shown in Fig. 12–13a. Construct the v9t and a–t graphs for 0 … t … 30 s. s (m)

150

s  6t  30

30

s  0.3t2

10

30

t (s)

(a)

Solution v–t Graph. Since v = ds>dt, the v9t graph can be determined by differentiating the equations defining the s–t graph, Fig. 12–13a. We have ds 0 … t 6 10 s; s = (0.3t 2) m v = = (0.6t) m>s dt ds 10 s 6 t … 30 s; s = (6t - 30) m v = = 6 m>s dt The results are plotted in Fig. 12–13b. We can also obtain specific values of v by measuring the slope of the s–t graph at a given instant. For example, at t = 20 s, the slope of the s–t graph is determined from the straight line from 10 s to 30 s, i.e., t = 20 s;

v =

v  0.6t

v6

6

30

10

t (s)

(b)

s 150 m - 30 m = = 6 m>s t 30 s - 10 s

a–t Graph. Since a = dv>dt, the a–t graph can be determined by differentiating the equations defining the lines of the v9t graph. This yields dv 0 … t 6 10 s; v = (0.6t) m>s a = = 0.6 m>s2 dt dv 10 6 t … 30 s; v = 6 m>s a = = 0 dt The results are plotted in Fig. 12–13c. NOTE: Show that a = 0.6 m>s2 when t = 5 s by measuring the slope

of the v9t graph.

v (m/s)

a (m/s2)

0.6

10

30 (c)

Fig. 12–13

t (s)

24

12

Chapter 12

ExamplE

K i n e m at i C s

partiCle

12.7 The car in Fig. 12–14a starts from rest and travels along a straight track such that it accelerates at 10 m>s2 for 10 s, and then decelerates at 2 m>s2. Draw the v9t and s–t graphs and determine the time t needed to stop the car. How far has the car traveled?

a (m/s2)

10 A1 t¿ A2

10

2

of a

t (s)

Solution v–t Graph. Since dv = a dt, the v9t graph is determined by integrating the straight-line segments of the a–t graph. Using the initial condition v = 0 when t = 0, we have t

v

10 dt, v = 10t L0 L0 When t = 10 s, v = 10(10) = 100 m>s. Using this as the initial condition for the next time period, we have 0 … t 6 10 s;

(a)

dv =

a = (10) m>s2;

10 s 6 t … t; a = (-2) m>s2;

t

v

L100 m>s

L10 s

dv =

-2 dt, v = ( -2t + 120) m>s

When t = t we require v = 0. This yields, Fig. 12–14b, t = 60 s

v (m/s) v  10t 100 v  2t  120

10

t¿  60

t (s)

(b)

Ans.

A more direct solution for t is possible by realizing that the area under the a–t graph is equal to the change in the car’s velocity. We require v = 0 = A1 + A2 , Fig. 12–14a. Thus 0 = 10 m>s2(10 s) + (-2 m>s2)(t - 10 s) t = 60 s Ans. s–t Graph. Since ds = v dt, integrating the equations of the v9t graph yields the corresponding equations of the s–t graph. Using the initial condition s = 0 when t = 0, we have s

t

10t dt, s = (5t2) m L0 L0 When t = 10 s, s = 5(10)2 = 500 m. Using this initial condition, 0 … t … 10 s;

v = (10t) m>s;

ds =

s

s (m) 3000 s  5t2 500

t

(-2t + 120) dt L500 m L10 s s - 500 = -t2 + 120t - [-(10)2 + 120(10)] s = (-t2 + 120t - 600) m When t = 60 s, the position is

10 s … t … 60 s; v = (-2t + 120) m>s;

s = -(60)2 + 120(60) - 600 = 3000 m The s–t graph is shown in Fig. 12–14c.

s  t2  120t  600 10

60 (c)

Fig. 12–14

ds =

t (s)

Ans.

NOTE: A direct solution for s is possible when t = 60 s, since the triangular area under the v9t graph would yield the displacement s = s - 0 from t = 0 to t = 60 s. Hence, s = 12(60 s)(100 m>s) = 3000 m Ans.

12.3

ExamplE

reCtilinear KinematiCs: erratiC motion

25

12.8

12

The v–s graph describing the motion of a motorcycle is shown in Fig. 12–15a. Construct the a–s graph of the motion and determine the time needed for the motorcycle to reach the position s = 160 m. Solution a–s Graph. Since the equations for segments of the v–s graph are given, the a–s graph can be determined using a ds = v dv. 0 … s 6 80 m; v = (0.2s + 4) m>s dv d = (0.2s + 4) (0.2s + 4) = 0.04s + 0.8 ds ds 80 m 6 s … 160 m; v = 20 m>s dv d a = v = (20) (20) = 0 ds ds The results are plotted in Fig. 12–15b. a = v

time. The time can be obtained using the v–s graph and v = ds>dt, because this equation relates v, s, and t. For the first segment of motion, s = 0 when t = 0, so ds ds 0 … s 6 80 m; v = (0.2s + 4) m>s; dt = = v 0.2s + 4 t s ds dt = L0 L0 0.2s + 4 0.2s + 4 t = c 5 ln a bd s 4 0.2(80) + 4 At s = 80 m, t = 5 ln c d = 8.047 s. Therefore, using these 4 initial conditions for the second segment of motion, 80 m 6 s … 160 m;

v = 20 m>s; t

dt = s

ds ds = v 20

ds ; 20 L8.047 s L80 m s s t - 8.047 = - 4; t = a + 4.047b s 20 20 Therefore, at s = 160 m, 160 t = + 4.047 = 12.0 s Ans. 20 dt =

NOTE: The graphical results can be checked in part by calculating slopes.

For example, at s = 0, a = v(dv>ds) = 4(20 - 4)>80 = 0.8 m>s2. Also, the results can be checked in part by inspection. The v–s graph indicates the initial increase in velocity (acceleration) followed by constant velocity (a = 0).

v (m/s)

20

v  0.2s  4 v  20

4 80

160

s (m)

(a)

a (m/s2) a  0.04s  0.8 4 0.8

a0 80

160 (b)

Fig. 12–15

s (m)

26

Chapter 12

K i n e m at i C s

of a

partiCle

preliMinary prObleM

12 P12–2.

e) Draw the v–t graph if v = 0 when t = 0. Find the equation v = f(t) for each segment.

a) Draw the s–t and a–t graphs if s = 0 when t = 0. v (m/s) v  2t

4

a (m/s2) 2 4

t (s)

2

2

t (s)

2

b) Draw the a–t and v–t graphs. s (m) 2

f) Determine v at s = 2 m if v = 1 m>s at s = 0.

s  2t  2

a (m/s)

t (s)

1

4

c) Draw the v–t and s–t graphs if v = 0, s = 0 when t = 0. a (m/s2) s (m)

2 2

t (s)

2

g) Determine a at s = 1 m. d) Determine s and a when t = 3 s if s = 0 when t = 0.

v (m/s)

v (m/s)

4

2

2

4

t (s)

Prob. P12–2

2

s (m)

12.3

27

reCtilinear KinematiCs: erratiC motion

FunDaMental prObleMs F12–9. The particle travels along a straight track such that its position is described by the s9t graph. Construct the v9t graph for the same time interval. s (m)

12 F12–12. The sports car travels along a straight road such that its acceleration is described by the graph. Construct the v9s graph for the same interval and specify the velocity of the car when s = 10 m and s = 15 m. a (m/s2) 10

s  108

108

s  0.5t3 0 6

8

10

5

t (s)

Prob. F12–9 F12–10. The sports car travels along a straight road such that its position is described by the graph. Construct the v9t and a-t graphs for the time interval 0 ≤ t ≤ 10 s. s (m)

10

s (m)

15

Prob. F12–12 F12–13. The dragster starts from rest and has an acceleration described by the graph. Construct the v9t graph for the time interval 0 … t … t, where t is the time for the car to come to rest. a (m/s2)

225 30t

s

20

75

75

t¿

0 s

3t2

0

5

10

t (s)

10

t (s)

5

Prob. F12–10 F12–11. A bicycle travels along a straight road where its velocity is described by the v9s graph. Construct the a9s graph for the same interval. v (m/s)

Prob. F12–13 F12–14. The dragster starts from rest and has a velocity described by the graph. Construct the s9t graph during the time interval 0 … t … 15 s. Also, determine the total distance traveled during this time interval. v (m/s)

v  30t

10

150 v  0.25s

v  15t  225

40

Prob. F12–11

s (m)

5

Prob. F12–14

15

t (s)

28

Chapter 12

K i n e m at i C s

of a

partiCle

prObleMs

12

12–35. A train starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the train decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the v9t graph and determine the maximum speed of the train. *12–36. If the position of a particle is defined by s = [2 sin (p>5)t + 4] m, where t is in seconds, construct the s9t, v9t, and a9t graphs for 0 … t … 10 s. 12–37. A particle starts from s = 0 and travels along a straight line with a velocity v = (t2 - 4t + 3) m>s, where t is in seconds. Construct the v–t and a–t graphs for the time interval 0 … t … 4 s.

12–41. The velocity of a car is plotted as shown. Determine the total distance the car moves until it stops (t = 80 s). Construct the a–t graph.

v (m/s)

10

12–38. Two rockets start from rest at the same elevation. Rocket A accelerates vertically at 20 m>s2 for 12 s and then maintains a constant speed. Rocket B accelerates at 15 m>s2 until reaching a constant speed of 150 m>s. Construct the a–t, v–t, and s–t graphs for each rocket until t = 20 s. What is the distance between the rockets when t = 20 s?

40

80

t (s)

Prob. 12–41

12–39. If the position of a particle is defined by s = [3 sin(p>4)t + 8] m, where t is in seconds, construct the s9t, v9t, and a9t graphs for 0 … t … 10 s. *12–40. The s–t graph for a train has been experimentally determined. From the data, construct the v–t and a–t graphs for the motion; 0 … t … 40 s. For 0 … t … 30 s, the curve is s = (0.4t2) m, and then it becomes straight for t Ú 30 s.

12–42. The snowmobile moves along a straight course according to the v–t graph. Construct the s–t and a–t graphs for the same 50-s time interval.When t = 2, s = 0.

s (m) v (m/s) 600 12 360

30

40

Prob. 12–40

t (s)

30

Prob. 12–42

50

t (s)

12.3 12–43. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure. The acceleration and deceleration that occur are constant and both have a magnitude of 4 m>s2. If the plates are spaced 200 mm apart, determine the maximum velocity vmax and the time t for the particle to travel from one plate to the other. Also draw the s–t graph. When t = t>2 the particle is at s = 100 mm. *12–44. The v–t graph for a particle moving through an electric field from one plate to another has the shape shown in the figure, where t = 0.2 s and vmax = 10 m>s. Draw the s–t and a–t graphs for the particle. When t = t>2 the particle is at s = 0.5 m.

29

reCtilinear KinematiCs: erratiC motion

12–46. A two-stage rocket is fired vertically from rest at s = 0 with the acceleration as shown. After 30 s the first 12 stage, A, burns out and the second stage, B, ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 60 s.

a (m/s2) B 24

A

smax 12 v s

vmax

30

t (s)

60

Prob. 12–46 t¿/2

t

t¿

Probs. 12–43/44

12–45. The motion of a jet plane just after landing on a runway is described by the a–t graph. Determine the time t when the jet plane stops. Construct the v–t and s–t graphs for the motion. Here s = 0, and v = 150 m>s when t = 0.

12–47. The a–t graph of the bullet train is shown. If the train starts from rest, determine the elapsed time t before it again comes to rest.What is the total distance traveled during this time interval? Construct the v–t and s–t graphs.

a (m/s2)

a (m/s2) 10

20

t¿

a

t (s)

0.1t

3

10

30

20

Prob. 12–45

1

( 15 )t

a 75

Prob. 12–47

5 t¿

t (s)

30

Chapter 12

K i n e m at i C s

of a

partiCle

*12–48. The v–t graph for a train has been experimentally 12 determined. From the data, construct the s–t and a–t graphs for the motion for 0 … t … 180 s. When t = 0, s = 0.

12–50. The jet car is originally traveling at a velocity of 10 m>s when it is subjected to the acceleration shown. Determine the car’s maximum velocity and the time t when it stops. When t = 0, s = 0.

a (m/s2)

v (m/s)

10 6 6 t¿ 15

t (s)

4 60

120

180

t (s)

Prob. 12–50

Prob. 12–48

12–49. The v–s graph for a go-cart traveling on a straight road is shown. Determine the acceleration of the go-cart at s = 50 m and s = 150 m. Draw the a–s graph.

12–51. The race car starts from rest and travels along a straight road until it reaches a speed of 26 m>s in 8 s as shown on the v–t graph. The flat part of the graph is caused by shifting gears. Draw the a–t graph and determine the maximum acceleration of the car.

v (m/s) 6

v (m/s) 26 v  4t  6

8

14 v  3.5t

100

200

Prob. 12–49

s (m)

4

5

Prob. 12–51

8

t (s)

12.3 *12–52. A car starts from rest and travels along a straight road with a velocity described by the graph. Determine the total distance traveled until the car stops. Construct the s–t and a–t graphs.

31

reCtilinear KinematiCs: erratiC motion

12–55. The v–t graph for the motion of a car as it moves along a straight road is shown. Draw the s–t and a–t graphs. 12 Also determine the average speed and the distance traveled for the 15-s time interval. When t = 0, s = 0.

v(m/s)

30

v

t

v

30

60

0.5t

45 t(s)

90

v (m/s) 15

Prob. 12–52

v  0.6t 2

12–53. The v–s graph for an airplane traveling on a straight runway is shown. Determine the acceleration of the plane at s = 100 m and s = 150 m. Draw the a–s graph. 5 v (m/s)

50 40

t (s)

Prob. 12–55

v

0.1s

30

*12–56. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the total distance the motorcycle travels until it stops when t = 15 s. Also plot the a–t and s–t graphs.

0.4s

v

15

100

s (m)

200

Prob. 12–53 12–54. The a–s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v–s graph. At s = 0, v = 0. a (m/s2)

12–57. A motorcycle starts from rest at s = 0 and travels along a straight road with the speed shown by the v–t graph. Determine the motorcycle’s acceleration and position when t = 8 s and t = 12 s.

v (m/s)

5

v  1.25t

v5

2

v  t  15

200

Prob. 12–54

300

s (m)

4

10

Probs. 12–56/57

15

t (s)

32

Chapter 12

K i n e m at i C s

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partiCle

12–58. Two cars start from rest side by side and travel 2 12 along a straight road. Car A accelerates at 4 m>s for 10 s and then maintains a constant speed. Car B accelerates at 5 m>s 2 until reaching a constant speed of 25 m>s and then maintains this speed. Construct the a–t, v–t, and s–t graphs for each car until t = 15 s. What is the distance between the two cars when t = 15 s?

12–62. The motion of a train is described by the a–s graph shown. Draw the v–s graph if v = 0 at s = 0.

a (m/s2)

12–59. A particle travels along a curve defined by the equation s = (t3 - 3t2 + 2t) m, where t is in seconds. Draw the s – t, v – t, and a – t graphs for the particle for 0 … t … 3 s.

3

300

*12–60. The speed of a train during the first minute has been recorded as follows: t 1s2

600

s (m)

Prob. 12–62

0 20 40 60

v 1m>s2

0 16 21 24

Plot the v–t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled.

12–61. A two-stage rocket is fired vertically from rest with the acceleration shown. After 15 s the first stage A burns out and the second stage B ignites. Plot the v–t and s–t graphs which describe the motion of the second stage for 0 … t … 40 s.

12–63. If the position of a particle is defined as s = (5t - 3t2) m, where t is in seconds, construct the s–t, v–t, and a–t graphs for 0 … t … 2.5 s.

*12–64. From experimental data, the motion of a jet plane while traveling along a runway is defined by the v–t graph. Construct the s–t and a–t graphs for the motion. When t = 0, s = 0.

a (m/s2) B v (m/s) A

60

20 15 20 15

40

Prob. 12–61

t (s)

5

20

Prob. 12–64

30

t (s)

12.3 12–65. The v–s graph for a test vehicle is shown. Determine its acceleration when s = 100 m and when s = 175 m.

33

reCtilinear KinematiCs: erratiC motion

12–67. The boat travels along a straight line with the speed described by the graph. Construct the s–t and a–s graphs. 12 Also, determine the time required for the boat to travel a distance s = 400 m if s = 0 when t = 0.

v (m/s)

v (m/s)

80 50 v  0.2s v2  4s 150

s (m)

200

20 s (m)

Prob. 12–65

100

400

Prob. 12–67

12–66. The a–t graph for a car is shown. Construct the v–t and s–t graphs if the car starts from rest at t = 0. At what time t does the car stop?

*12–68. The a–s graph for a train traveling along a straight track is given for the first 400 m of its motion. Plot the v–s graph. v = 0 at s = 0.

a (m/s2)

a (m/s2)

5

2 t¿ 10 2

t (s) 200

Prob. 12–66

Prob. 12–68

400

s (m)

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12.4

12

General Curvilinear Motion

Curvilinear motion occurs when a particle moves along a curved path. Since this path is often described in three dimensions, vector analysis will be used to formulate the particle’s position, velocity, and acceleration.* In this section the general aspects of curvilinear motion are discussed, and in subsequent sections we will consider three types of coordinate systems often used to analyze this motion.

Position.

s s

r O Position

Path

(a)

s

s

Consider a particle located at a point on a space curve defined by the path function s(t), Fig. 12–16a. The position of the particle, measured from a fixed point O, will be designated by the position vector r = r(t). Notice that both the magnitude and direction of this vector will change as the particle moves along the curve.

Displacement.

Suppose that during a small time interval t the particle moves a distance s along the curve to a new position, defined by r = r + r, Fig. 12–16b. The displacement r represents the change in the particle’s position and is determined by vector subtraction; i.e., r = r - r.

Velocity. During the time t, the average velocity of the particle is

r r¿

vavg =

r O Displacement (b)

The instantaneous velocity is determined from this equation by letting t S 0, and consequently the direction of r approaches the tangent to the curve. Hence, v = lim (r> t) or S t

0

v = v s

r

O

Velocity (c)

Fig. 12–16

r t

dr dt

(12–7)

Since dr will be tangent to the curve, the direction of v is also tangent to the curve, Fig. 12–16c. The magnitude of v, which is called the speed, is obtained by realizing that the length of the straight line segment r in Fig. 12–16b approaches the arc length s as t S 0, we have v = lim (r> t) = lim (s> t), or S S t

0

t

0

v =

ds dt

(12–8)

Thus, the speed can be obtained by differentiating the path function s with respect to time. *A summary of some of the important concepts of vector analysis is given in Appendix B.

12.4

Acceleration.

12

v t

(d)

where v = v - v. To study this time rate of change, the two velocity vectors in Fig. 12–16d are plotted in Fig. 12–16e such that their tails are located at the fixed point O and their arrowheads touch points on a curve. This curve is called a hodograph, and when constructed, it describes the locus of points for the arrowhead of the velocity vector in the same manner as the path s describes the locus of points for the arrowhead of the position vector, Fig. 12–16a. To obtain the instantaneous acceleration, let t S 0 in the above equation. In the limit v will approach the tangent to the hodograph, and so a = lim (v> t), or S t

v

v¿

If the particle has a velocity v at time t and a velocity v = v + v at t + t, Fig. 12–16d, then the average acceleration of the particle during the time interval t is

aavg =

35

General Curvilinear motion

v

v v¿

O¿

(e)

0

a =

dv dt

(12–9)

Hodograph v a O¿

Substituting Eq. 12–7 into this result, we can also write

a =

(f)

d2r dt2

By definition of the derivative, a acts tangent to the hodograph, Fig. 12–16f, and, in general it is not tangent to the path of motion, Fig. 12–16g. To clarify this point, realize that v and consequently a must account for the change made in both the magnitude and direction of the velocity v as the particle moves from one point to the next along the path, Fig. 12–16d. However, in order for the particle to follow any curved path, the directional change always “swings” the velocity vector toward the “inside” or “concave side” of the path, and therefore a cannot remain tangent to the path. In summary, v is always tangent to the path and a is always tangent to the hodograph.

a Acceleration (g)

Fig. 12–16

path

36

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12.5

12

Curvilinear Motion: Rectangular Components

Occasionally the motion of a particle can best be described along a path that can be expressed in terms of its x, y, z coordinates.

Position.

If the particle is at point (x, y, z) on the curved path s shown in Fig. 12–17a, then its location is defined by the position vector

z

s k i

r = xi + yj + zk

z r  xi  yj  zk

j y

x

y x

Position

(12–10)

When the particle moves, the x, y, z components of r will be functions of time; i.e., x = x(t), y = y(t), z = z(t), so that r = r(t). At any instant the magnitude of r is defined from Eq. B–3 in Appendix B as

(a)

r = 2x2 + y2 + z2 And the direction of r is specified by the unit vector ur = r>r.

z

Velocity. The first time derivative of r yields the velocity of the particle. Hence, s

v =

v  vxi  vyj  vzk y x Velocity

dr d d d = (xi) + (yj) + (zk) dt dt dt dt

When taking this derivative, it is necessary to account for changes in both the magnitude and direction of each of the vector’s components. For example, the derivative of the i component of r is

(b)

d dx di (xi) = i + x dt dt dt Fig. 12–17

The second term on the right side is zero, provided the x, y, z reference frame is fixed, and therefore the direction (and the magnitude) of i does not change with time. Differentiation of the j and k components may be carried out in a similar manner, which yields the final result, v =

dr = vxi + vy j + vzk dt

(12–11)

where # # # vx = x vy = y vz = z

(12–12)

12.5

Curvilinear motion: reCtanGular Components

37

# # # The “dot” notation x, y, z represents the first time derivatives of x = x(t), y = y(t), z = z(t), respectively. The velocity has a magnitude that is found from

12

v = 2v2x + v2y + v2z and a direction that is specified by the unit vector uv = v>v. As discussed in Sec. 12.4, this direction is always tangent to the path, as shown in Fig. 12–17b. z

Acceleration. The acceleration of the particle is obtained by taking the first time derivative of Eq. 12–11 (or the second time derivative of Eq. 12–10). We have

s a  axi  ayj  azk

x

dv a = = ax i + ay j + az k dt

(12–13)

# $ ax = vx = x # $ ay = vy = y # $ az = vz = z

(12–14)

where

Here ax , ay , az represent, respectively, the first time derivatives of vx = vx(t), vy = vy(t), vz = vz(t), or the second time derivatives of the functions x = x(t), y = y(t), z = z(t). The acceleration has a magnitude

a = 2a2x + a2y + a2z and a direction specified by the unit vector ua = a>a. Since a represents the time rate of change in both the magnitude and direction of the velocity, in general a will not be tangent to the path, Fig. 12–17c.

Acceleration (c)

y

38

12

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important points • Curvilinear motion can cause changes in both the magnitude and direction of the position, velocity, and acceleration vectors.

• The velocity vector is always directed tangent to the path. • In general, the acceleration vector is not tangent to the path, but rather, it is tangent to the hodograph.

• If the motion is described using rectangular coordinates, then the components along each of the axes do not change direction, only their magnitude and sense (algebraic sign) will change.

• By considering the component motions, the change in magnitude and direction of the particle’s position and velocity are automatically taken into account.

procedure for analysis Coordinate System. • A rectangular coordinate system can be used to solve problems for which the motion can conveniently be expressed in terms of its x, y, z components. Kinematic Quantities. • Since rectilinear motion occurs along each coordinate axis, the motion along each axis is found using v = ds>dt and a = dv>dt; or in cases where the motion is not expressed as a function of time, the equation a ds = v dv can be used.

• In two dimensions, the equation of the path y = f (x) can be used

to relate the x and y components of velocity and acceleration by applying the chain rule of calculus. A review of this concept is given in Appendix C.

• Once the x, y, z components of v and a have been determined, the magnitudes of these vectors are found from the Pythagorean theorem, Eq. B–3, and their coordinate direction angles from the components of their unit vectors, Eqs. B–4 and B–5.

12.5

ExamplE

12.9

12

At any instant the horizontal position of the weather balloon in Fig.  12–18a is defined by x = (2t) m, where t is in seconds. If the equation of the path is y = x2 >5, determine the magnitude and direction of the velocity and the acceleration when t = 2 s.

y

B

y

Solution velocity.

39

Curvilinear motion: reCtanGular Components

x2 5

The velocity component in the x direction is

x

A

d # vx = x = (2t) = 2 m>s S dt

4m

To find the relationship between the velocity components we will use the chain rule of calculus. When t = 2 s, x = 2122 = 4 m, Fig. 12–18a, and so

(a)

d 2 # # vy = y = (x >5) = 2xx >5 = 2(4)(2)>5 = 3.20 m>s c dt

When t = 2 s, the magnitude of velocity is therefore

v = 2(2 m>s)2 + (3.20 m>s)2 = 3.77 m>s

Ans.

The direction is tangent to the path, Fig. 12–18b, where uv = tan

-1

vy vx

= tan

-1 3.20

2

v  3.77 m/s

= 58.0

Ans.

uv  58.0 B

acceleration. The relationship between the acceleration components is determined using the chain rule. (See Appendix C.) We have

(b)

d # ax = vx = (2) = 0 dt d # # # # $ ay = vy = (2xx >5) = 2(x)x >5 + 2x(x)>5 dt Thus,

= 2(2)2 >5 + 2(4)(0)>5 = 1.60 m>s2 c a = 2(0)2 + (1.60)2 = 1.60 m>s2

Ans.

a  1.60 m/s2 ua  90 B

The direction of a, as shown in Fig. 12–18c, is

(c)

1.60 ua = tan-1 = 90 0

Ans.

NOTE: It is also possible to obtain vy and ay by first expressing

y = f (t) = (2t)2 >5 = 0.8t2 and then taking successive time derivatives.

Fig. 12–18

40

12

Chapter 12

ExamplE

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12.10 For a short time, the path of the plane in Fig. 12–19a is described by y = (0.001x2) m. If the plane is rising with a constant upward velocity of 10 m>s, determine the magnitudes of the velocity and acceleration of the plane when it reaches an altitude of y = 100 m.

y

x

Solution When y = 100 m, then 100 = 0.001x2 or x = 316.2 m. Also, due to constant velocity vy = 10 m>s, so y = vy t;

100 m = (10 m>s) t

t = 10 s

velocity. Using the chain rule (see Appendix C) to find the relationship between the velocity components, we have y = 0.001x2 d # # vy = y = (0.001x 2) = (0.002x)x = 0.002 xvx dt

y

(1)

Thus y  0.001x2

10 m>s = 0.002(316.2 m)(vx) vx = 15.81 m>s

100 m x (a)

The magnitude of the velocity is therefore v = 2v2x + v2y = 2(15.81 m>s)2 + (10 m>s)2 = 18.7 m>s

Ans.

acceleration. Using the chain rule, the time derivative of Eq. (1) gives the relation between the acceleration components. # # # $ ay = vy = (0.002x)x + 0.002x(x) = 0.002(vx2 + xax) y

# When x = 316.2 m, vx = 15.81 m>s , vy = ay = 0, vy a

100 m

0 = 0.002 3 (15.81 m>s)2 + 316.2 m(ax)4 ax = -0.791 m>s2

v vx x

(b)

Fig. 12–19

The magnitude of the plane’s acceleration is therefore a = 2a2x + a2y = 2(-0.791 m>s2)2 + (0 m>s2)2 = 0.791 m>s2

These results are shown in Fig. 12–19b.

Ans.

12.6

12.6

motion of a projeCtile

41

Motion of a Projectile

12

The free-flight motion of a projectile is often studied in terms of its rectangular components. To illustrate the kinematic analysis, consider a projectile launched at point (x0 , y0), with an initial velocity of v0 , having components (v0)x and (v0)y , Fig. 12–20. When air resistance is neglected, the only force acting on the projectile is its weight, which causes the projectile to have a constant downward acceleration of approximately ac = g = 9.81 m>s2.* y

ag

vx v0

(v0)y

vy

(v0)x

v

r

y y0 x x0 x

Fig. 12–20

Horizontal Motion.

Since ax = 0, application of the constant acceleration equations, 12–4 to 12–6, yields + ) (S + ) (S + ) (S

v = v0 + act x = x0 + v0t + 12 act2; v2 = v20 + 2ac(x - x0);

vx = (v0)x x = x0 + (v0)xt vx = (v0)x

The first and last equations indicate that the horizontal component of velocity always remains constant during the motion.

Vertical Motion.

Since the positive y axis is directed upward, then ay = -g. Applying Eqs. 12–4 to 12–6, we get (+ c ) (+ c ) (+ c )

v = v0 + act; y = y0 + v0t + 12 act2; v2 = v20 + 2ac(y - y0);

vy = (v0)y - gt y = y0 + (v0)yt - 12 gt2 v2y = (v0)2y - 2g(y - y0)

Recall that the last equation can be formulated on the basis of eliminating the time t from the first two equations, and therefore only two of the above three equations are independent of one another. *This assumes that the earth’s gravitational field does not vary with altitude.

Each picture in this sequence is taken after the same time interval. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity when released. Both balls accelerate downward at the same rate, and so they remain at the same elevation at any instant. This acceleration causes the difference in elevation between the balls to increase between successive photos. Also, note the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction remains constant.

42

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To summarize, problems involving the motion of a projectile can have at most three unknowns since only three independent equations can be written; that is, one equation in the horizontal direction and two in the vertical direction. Once vx and vy are obtained, the resultant velocity v, which is always tangent to the path, can be determined by the vector sum as shown in Fig. 12–20.

12

procedure for analysis Once thrown, the basketball follows a parabolic trajectory.

Coordinate System. • Establish the fixed x, y coordinate axes and sketch the trajectory of the particle. Between any two points on the path specify the given problem data and identify the three unknowns. In all cases the acceleration of gravity acts downward and equals 9.81 m>s2. The particle’s initial and final velocities should be represented in terms of their x and y components.

• Remember that positive and negative position, velocity, and

acceleration components always act in accordance with their associated coordinate directions.

Kinematic Equations. • Depending upon the known data and what is to be determined, a choice should be made as to which three of the following four equations should be applied between the two points on the path to obtain the most direct solution to the problem. Horizontal motion. • The velocity in the horizontal or x direction is constant, i.e., vx = (v0)x , and x = x0 + (v0)x t Vertical motion. • In the vertical or y direction only two of the following three equations can be used for solution. vy = (v0)y + ac t y = y0 + (v0)y t + 12 ac t2 Gravel falling off the end of this conveyor belt follows a path that can be predicted using the equations of constant acceleration. In this way the location of the accumulated pile can be determined. Rectangular coordinates are used for the analysis since the acceleration is only in the vertical direction.

v2y = (v0)2y + 2ac( y - y0) For example, if the particle’s final velocity vy is not needed, then the first and third of these equations will not be useful.

12.6

ExamplE

12.11

43

12

A sack slides off the ramp, shown in Fig. 12–21, with a horizontal velocity of 12 m>s. If the height of the ramp is 6 m from the floor, determine the time needed for the sack to strike the floor and the range R where sacks begin to pile up. y A

12 m/s

x ag

6m

B

C

R

Fig. 12–21

Solution Coordinate system. The origin of coordinates is established at the beginning of the path, point A, Fig. 12–21. The initial velocity of a sack has components (vA)x = 12 m>s and (vA)y = 0. Also, between points A and B the acceleration is ay = -9.81 m>s2. Since (vB)x = (vA)x = 12 m>s, the three unknowns are (vB)y , R, and the time of flight tAB . Here we do not need to determine (vB)y . vertical Motion. The vertical distance from A to B is known, and therefore we can obtain a direct solution for tAB by using the equation (+ c ) yB = yA + (vA)ytAB + 12 act2AB -6 m = 0 + 0 + 12(-9.81 m>s2)t2AB tAB = 1.11 s horizontal Motion. as follows: + ) (S

motion of a projeCtile

Ans.

Since tAB has been calculated, R is determined

xB = xA + (vA)xtAB R = 0 + 12 m>s (1.11 s) R = 13.3 m Ans. NOTE: The calculation for tAB also indicates that if a sack were released from rest at A, it would take the same amount of time to strike the floor at C, Fig. 12–21.

44

12

Chapter 12

ExamplE

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12.12 The chipping machine is designed to eject wood chips at vO = 7.5 m>s as shown in Fig. 12–22. If the tube is oriented at 30° from the horizontal, determine how high, h, the chips strike the pile if at this instant they land on the pile 6 m from the tube. y vO  7.5 m/s 30

x

A

O 1.2 m

h

6m

Fig. 12–22

Solution Coordinate system. When the motion is analyzed between points O and A, the three unknowns are the height h, time of flight tOA , and vertical component of velocity (vA)y . [Note that (vA)x = (vO)x .] With the origin of coordinates at O, Fig. 12–22, the initial velocity of a chip has components of (vO)x = (7.5 cos 30) m>s = 6.50 m>s S (vO)y = (7.5 sin 30) m>s = 3.75 m>s c Also, (vA)x = (vO)x = 6.50 m>s and ay = -9.81 m>s2. Since we do not need to determine (vA)y , we have horizontal Motion. + ) (S

xA = xO + (vO)xtOA 6 m = 0 + (6.50 m>s)tOA tOA = 0.923 s

vertical Motion. Relating tOA to the initial and final elevations of a chip, we have ( + c ) yA = yO + (vO)ytOA + 12 ac t2OA (h - 1.2 m) = 0 + (3.75 m>s)(0.923 s) + 12( -9.81 m>s2)(0.923 s)2 h = 0.483 m Ans. NOTE: We can determine (vA)y by using (vA)y = (vO)y + actOA .

12.6

ExamplE

45

motion of a projeCtile

12.13

12

The track for this racing event was designed so that riders jump off the slope at 30°, from a height of 1 m. During a race it was observed that the rider shown in Fig. 12–23a remained in mid air for 1.5 s. Determine the speed at which he was traveling off the ramp, the horizontal distance he travels before striking the ground, and the maximum height he attains. Neglect the size of the bike and rider.

(a)

Solution Coordinate system. As shown in Fig. 12–23b, the origin of the coordinates is established at A. Between the end points of the path AB the three unknowns are the initial speed vA , range R, and the vertical component of velocity (vB)y .

y

30 A

C

vertical Motion. Since the time of flight and the vertical distance between the ends of the path are known, we can determine vA . (+ c )

yB = yA + (vA)ytAB +

1 2 2 actAB

x

h

1m R

(b)

m>s )(1.5 s) -1 m = 0 + vA sin 30(1.5 s) + Ans. vA = 13.38 m>s = 13.4 m>s horizontal Motion. The range R can now be determined. + ) (S xB = xA + (vA)xtAB 1 2 (-9.81

2

2

R = 0 + 13.38 cos 30 m>s (1.5 s) = 17.4 m Ans. In order to find the maximum height h we will consider the path AC, Fig. 12–23b. Here the three unknowns are the time of flight tAC , the horizontal distance from A to C, and the height h. At the maximum height (vC)y = 0, and since vA is known, we can determine h directly without considering tAC using the following equation. (vC)2y = (vA)2y + 2ac[ yC - yA] 02 = (13.38 sin 30 m>s)2 + 2(-9.81 m>s2)[(h - 1 m) - 0] Ans. h = 3.28 m NOTE: Show that the bike will strike the ground at B with a velocity having components of (vB)x = 11.6 m>s S , (vB)y = 8.02 m>s T

Fig. 12–23

B

46

Chapter 12

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preliMinary prObleMs

12

P12–3. Use the chain-rule and find y· and y¨ in terms of

x, x· and x¨ if

The particle travels from A to B. Identify the three unknowns, and write the three equations needed to solve for them. P12–5.

a) y = 4x2

10 m/s

b) y = 3ex

30 A 8m

c) y = 6 sin x

B

x

Prob. P12–5

The particle travels from A to B. Identify the three unknowns, and write the three equations needed to solve for them. P12–4.

The particle travels from A to B. Identify the three unknowns, and write the three equations needed to solve for them. P12–6.

y

y 40 m/s A B

60 m/s B 20 m

Prob. P12–4

20

x A

x tAB  5 s

Prob. P12–6

12.6

47

motion of a projeCtile

FunDaMental prObleMs F12–15. If the x and y components of a particle’s velocity are vx = (32t) m>s and vy = 8 m>s, determine the equation of the path y = f(x), if x = 0 and y = 0 when t = 0. F12–16. A particle is traveling along the straight path. If its position along the x axis is x = (8t) m, where t is in seconds, determine its speed when t = 2 s.

12 F12–18. A particle travels along a straight-line path y = 0.5x. If the x component of the particle’s velocity is vx = (2t2) m>s, where t is in seconds, determine the magnitude of the particle’s velocity and acceleration when t = 4 s. y

y y  0.5x x

y  0.75x

Prob. F12–18 3m

x x  8t

F12–19. A particle is traveling along the parabolic path y = 0.25x2. If x = 8 m, vx = 8 m>s, and ax = 4 m>s2 when t = 2 s, determine the magnitude of the particle’s velocity and acceleration at this instant. y

4m

Prob. F12–16 y  0.25x2

F12–17. A particle is constrained to travel along the path. If x = (4t4) m, where t is in seconds, determine the magnitude of the particle’s velocity and acceleration when t = 0.5 s.

y

x

Prob. F12–19 F12–20. The box slides down the slope described by the equation y = (0.05x2) m, where x is in meters. If the box has x components of velocity and acceleration of vx = - 3 m>s and ax = -1.5 m>s2 at x = 5 m, determine the y components of the velocity and the acceleration of the box at this instant.

y2  4x

y x

y  0.05x2 x

x  (4t4) m

Prob. F12–17

Prob. F12–20

48

Chapter 12

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F12–21. The ball is kicked from point A with the initial 12 velocity vA = 10 m>s. Determine the maximum height h it reaches.

F12–25. A ball is thrown from A. If it is required to clear the wall at B, determine the minimum magnitude of its initial velocity vA.

F12–22. The ball is kicked from point A with the initial velocity vA = 10 m>s. Determine the range R, and the speed when the ball strikes the ground. y

y

B

xB B

vA  10 m/s A

vA h

30

C

30

A

2.4 m

x

x 0.9 m

x

Prob. F12–21/22 3.6 m

F12–23. Determine the speed at which the basketball at A must be thrown at the angle of 30 so that it makes it to the basket at B. y

B vA 30

A

Prob. F12–25

x

3m

1.5 m

F12–26. A projectile is fired with an initial velocity of vA = 150 m>s off the roof of the building. Determine the range R where it strikes the ground at B.

10 m

Prob. F12–23 F12–24. Water is sprayed at an angle of 90 from the slope at 20 m>s. Determine the range R.

y vA  150 m/s

vB  20 m/s A

5 4

5 3 4

x

150 m 3

R

B R

Prob. F12–24

Prob. F12–26

12.6

49

motion of a projeCtile

prObleMs

12

12–69. The velocity of a particle is given by v = 5 16t 2 i + 4t 3j + (5t + 2)k 6 m>s, where t is in seconds. If the particle is at the origin when t = 0, determine the magnitude of the particle’s acceleration when t = 2 s. Also, what is the x, y, z coordinate position of the particle at this instant?

12–74. A particle is traveling with a velocity of 2 v = 5 32te-0.2t i + 4e-0.8t j 6 m>s, where t is in seconds. Determine the magnitude of the particle’s displacement from t = 0 to t = 3 s. Use Simpson’s rule with n = 100 to evaluate the integrals. What is the magnitude of the particle’s acceleration when t = 2 s?

12–70. The position of a particle is defined by r = {5(cos 2t)i + 4(sin 2t)j} m, where t is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particle when t = 1 s. Also, prove that the path of the particle is elliptical.

12–75. A particle travels along the curve from A to B in 2 s. It takes 4 s for it to go from B to C and then 3 s to go from C to D. Determine its average speed when it goes from A to D.

y

12–71. The velocity of a particle is v = 5 3i + (6 - 2t)j 6 m>s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = 1 s to t = 3 s.

D B

y (y

40)2

15 m C

*12–72. If the velocity of a particle is defined as v(t) = {0.8t2i + 12t1 > 2j + 5k} m >s, determine the magnitude and coordinate direction angles a, b, g of the particle’s acceleration when t = 2 s. 12–73. When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path (y - 40)2 = 160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy = 180 m/s, determine the magnitudes of the rocket’s velocity and acceleration when it reaches an altitude of 80 m.

5m

10 m x

A

Prob. 12–75

*12–76. A particle travels along the curve from A to B in 5 s. It takes 8 s for it to go from B to C and then 10 s to go from C to A. Determine its average speed when it goes around the closed path.

160x

y

B

20 m

40 m x

Prob. 12–73

A

30 m

Prob. 12–76

C

x

50

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12–77. The position of a crate sliding down a ramp is given > 12 by x = (0.25t3) m, y = (1.5t2) m, z = (6 − 0.75t5 2) m, where t is in seconds. Determine the magnitude of the crate’s velocity and acceleration when t = 2 s.

12–81. A particle travels along the curve from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C.

12–78. A rocket is fired from rest at x = 0 and travels along a parabolic trajectory described by y2 = [120(103)x] m. 1 If the x component of acceleration is ax = a t2 b   m>s2, 4 where t  is in seconds, determine the magnitude of the rocket’s velocity and acceleration when t = 10 s.

y

12–79. The particle travels along the path defined by the parabola y = 0.5x 2. If the component of velocity along the x axis is vx = (5t) m>s, where t is in seconds, determine the particle’s distance from the origin O and the magnitude of its acceleration when t = 1 s. When t = 0, x = 0, y = 0.

30

C

45 30 m

B x

A y

Prob. 12–81

y  0.5x2

12–82. The roller coaster car travels down the helical path at constant speed such that the parametric equations that define its position are x = c sin kt, y = c cos kt, z = h − bt, where c, h, and b are constants. Determine the magnitudes of its velocity and acceleration.

x

O

Prob. 12–79

z

*12–80. The motorcycle travels with constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve.

y v0

p x) y  c sin ( –– L x

c L

Prob. 12–80

y

c L

x

Prob. 12–82

12.6 12–83. The flight path of the helicopter as it takes off from A is defined by the parametric equations x = (2t2) m and y = (0.04t3) m, where t is the time in seconds. Determine the distance the helicopter is from point A and the magnitudes of its velocity and acceleration when t = 10 s.

51

motion of a projeCtile

12–86. Neglecting the size of the ball, determine the magnitude vA of the basketball’s initial velocity and its 12 velocity when it passes through the basket.

B

y

30 A

vA 3m

2m 10 m

Prob. 12–86 A

x

Prob. 12–83 *12–84. Pegs A and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10 m>s, determine the magnitude of the velocity and acceleration of peg A when x = 1 m.

12–87. A projectile is fired from the platform at B. The shooter fires his gun from point A at an angle of 30°. Determine the muzzle speed of the bullet if it hits the projectile at C.

B C

y vA A A C B

D

10 m

30 1.8 m 20 m

x

v  10 m/s

Prob. 12–87

2

x  y2  1 4

*12–88. Determine the minimum initial velocity v0 and the corresponding angle u0 at which the ball must be kicked in order for it to just cross over the 3-m high fence.

Prob. 12–84 12–85. It is observed that the time for the ball to strike the ground at B is 2.5 s. Determine the speed vA and angle uA at which the ball was thrown.

v0

vA A

uA

3m

u0

1.2 m 50 m

Prob. 12–85

B

6m

Prob. 12–88

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12–89. A projectile is given a velocity v0 at an angle f above 12 the horizontal. Determine the distance d to where it strikes the sloped ground. The acceleration due to gravity is g. 12–90. A projectile is given a velocity v0. Determine the angle f at which it should be launched so that d is a maximum. The acceleration due to gravity is g.

12–93. The boy at A attempts to throw a ball over the roof of a barn with an initial speed of vA = 15 m>s. Determine the angle uA at which the ball must be thrown so that it reaches its maximum height at C. Also, find the distance d where the boy should stand to make the throw.

12–94. The boy at A attempts to throw a ball over the roof of a barn such that it is launched at an angle uA = 40°. Determine the minimum speed vA at which he must throw the ball so that it reaches its maximum height at C. Also, find the distance d where the boy must stand so that he can make the throw.

y

v0

φ

θ

x

C

d

Probs. 12–89/90 8m

vA

*12–92. Show that the girl at A can throw the ball to the boy at B by launching it at equal angles measured up or down from a 45° inclination. If vA = 10 m >s, determine the range R if this value is 15°, i.e., u1 = 45° − 15° = 30° and u2 = 45° + 15° = 60°. Assume the ball is caught at the same elevation from which it is thrown.

uA

A

12–91. The girl at A can throw a ball at vA = 10 m>s. Calculate the maximum possible range R = Rmax and the associated angle u at which it should be thrown. Assume the ball is caught at B at the same elevation from which it is thrown.

1m d

4m

Probs. 12–93/94

12–95. Measurements of a shot recorded on a videotape during a basketball game are shown. The ball passed through the hoop even though it barely cleared the hands of the player B who attempted to block it. Neglecting the size of the ball, determine the magnitude vA of its initial velocity and the height h of the ball when it passes over player B.

vA  10 m/s

C 30

u A

B

R

Probs. 12–91/92

A

vA

B h

2.1 m 7.5 m

Prob. 12–95

1.5 m

3m

12.6 *12–96. The golf ball is hit at A with a speed of vA = 40 m>s and directed at an angle of 30° with the horizontal as shown. Determine the distance d where the ball strikes the slope at B.

53

motion of a projeCtile

12–99. The projectile is launched with a velocity v0. Determine the range R, the maximum height h attained, 12 and the time of flight. Express the results in terms of the angle u and v0. The acceleration due to gravity is g.

B vA

y

40 m/s 30

5

1

d

A

v0

Prob. 12–96

h

u

x R

Prob. 12–99 12–97. It is observed that the skier leaves the ramp A at an  angle uA = 25 with the horizontal. If he strikes the ground at B, determine his initial speed vA and the time of flight tAB .

12–98. It is observed that the skier leaves the ramp A at an angle uA = 25 with the horizontal. If he strikes the ground at B, determine his initial speed vA and the speed at which he strikes the ground.

*12–100. The missile at A takes off from rest and rises vertically to B, where its fuel runs out in 8 s. If the acceleration varies with time as shown, determine the missile’s height hB and speed vB. If by internal controls the missile is then suddenly pointed 45° as shown, and allowed to travel in free flight, determine the maximum height attained, hC, and the range R to where it crashes at D.

45

C

B

vA

uA A

vB

hC hB

4m

3

A

5

D R

4

100 m

a (m/s2) 40 B 8

Probs. 12–97/98

t (s)

Prob. 12–100

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12–101. The velocity of the water jet discharging from the 12 orifice can be obtained from v = 22 gh, where h = 2 m is the depth of the orifice from the free water surface. Determine the time for a particle of water leaving the orifice to reach point B and the horizontal distance x where it hits the surface.

2m

A

12–103. If the dart is thrown with a speed of 10 m>s, determine the shortest possible time before it strikes the target. Also, what is the corresponding angle uA at which it should be thrown, and what is the velocity of the dart when it strikes the target? *12–104. If the dart is thrown with a speed of 10 m>s, determine the longest possible time when it strikes the target. Also, what is the corresponding angle uA at which it should be thrown, and what is the velocity of the dart when it strikes the target?

vA

4m 1.5 m A vA

B x

uA B

Prob. 12–101

Probs. 12–103/104

12–102. The man at A wishes to throw two darts at the target at B so that they arrive at the same time. If each dart is thrown with a speed of 10 m>s, determine the angles uC and uD at which they should be thrown and the time between each throw. Note that the first dart must be thrown at uC (7 uD), then the second dart is thrown at uD.

12–105. The drinking fountain is designed such that the nozzle is located from the edge of the basin as shown. Determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C.

vA 40 A

5m uC A

uD

50 mm

C

B

D B

Prob. 12–102

C

250 mm 100 mm

Prob. 12–105

12–106. The balloon A is ascending at the rate vA = 12 km>h and is being carried horizontally by the wind at vw = 20 km>h. If a ballast bag is dropped from the balloon at the instant h = 50 m, determine the time needed for it to strike the ground. Assume that the bag was released from the balloon with the same velocity as the balloon. Also, with what speed does the bag strike the ground?

vA

55

motion of a projeCtile

12.6

*12–108. A boy throws a ball at O in the air with a speed v0 at an angle u1. If he then throws another ball with the 12 same speed v0 at an angle u2 6 u1, determine the time between the throws so that the balls collide in midair at B.

12 km/h

vw

B

20 km/h

A

O

u1

y

u2

x

Prob. 12–108 h

Prob. 12–106

12–107. The snowmobile is traveling at 10 m>s when it leaves the embankment at A. Determine the time of flight from A to B and the range R of the trajectory.

A

12–109. Small packages traveling on the conveyor belt fall off into a l-m-long loading car. If the conveyor is running at a constant speed of vC = 2 m>s, determine the smallest and largest distance R at which the end A of the car may be placed from the conveyor so that the packages enter the car.

40

vc

2 m/s

30 3

5 4

3m

B

A

B

R R

Prob. 12–107

Prob. 12–109

1m

56

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12.7

12

When the path along which a particle travels is known, then it is often convenient to describe the motion using n and t coordinate axes which act normal and tangent to the path, respectively, and at the instant considered have their origin located at the particle.

O¿ n

O s

Curvilinear Motion: Normal and Tangential Components

un

Planar Motion.

ut t

Position (a) O¿ O¿ r r

ds

r

r

ds

r r

O¿

ds Radius of curvature (b)

Velocity. Since the particle moves, s is a function of time. As indicated in Sec. 12.4, the particle’s velocity v has a direction that is always tangent to the path, Fig. 12–24c, and a magnitude that is determined by taking the time derivative of the path function s = s(t), i.e., v = ds>dt (Eq. 12–8). Hence

O¿

r

r

v Velocity (c)

Consider the particle shown in Fig. 12–24a, which moves in a plane along a fixed curve, such that at a given instant it is at position s, measured from point O. We will now consider a coordinate system that has its origin on the curve, and at the instant considered this origin happens to coincide with the location of the particle. The t axis is tangent to the curve at the point and is positive in the direction of increasing s. We will designate this positive direction with the unit vector ut . A unique choice for the normal axis can be made by noting that geometrically the curve is constructed from a series of differential arc segments ds, Fig. 12–24b. Each segment ds is formed from the arc of an associated circle having a radius of curvature r (rho) and center of curvature O. The normal axis n is perpendicular to the t axis with its positive sense directed toward the center of curvature O, Fig. 12–24a. This positive direction, which is always on the concave side of the curve, will be designated by the unit vector un . The plane which contains the n and t axes is referred to as the embracing or osculating plane, and in this case it is fixed in the plane of motion.*

v = vut

(12–15)

# v = s

(12–16)

where

Fig. 12–24

*The osculating plane may also be defined as the plane which has the greatest contact with the curve at a point. It is the limiting position of a plane contacting both the point and the arc segment ds. As noted above, the osculating plane is always coincident with a plane curve; however, each point on a three-dimensional curve has a unique osculating plane.

12.7

57

Curvilinear motion: normal and tanGential Components

Acceleration. The acceleration of the particle is the time rate of change of the velocity. Thus, # # # a = v = vut + vut

12

O¿

du r

(12–17)

r

un

# In order to determine the time derivative ut , note that as the particle moves along the arc ds in time dt, ut preserves its magnitude of unity; however, its direction changes, and becomes ut= , Fig. 12–24d. As shown in Fig. 12–24e, we require ut= = ut + dut . Here dut stretches between the arrowheads of ut and ut= , which lie on an infinitesimal arc of radius ut = 1. Hence, dut has a magnitude of dut = (1) du, and its direction is defined by un . Consequently, dut = duun , and therefore #the time derivative becomes # # # ut = uun . Since ds = rdu, Fig. 12–24d, then u = s >r, and therefore

ds

u¿t ut

(d)

# # s v # ut = uun = un = un r r

Substituting into Eq. 12–17, a can be written as the sum of its two components,

a = atut + anun

un du u¿ t

(12–18)

dut

ut (e)

where # at = v

or

at ds = v dv

(12–19)

and

O¿

an =

v2 r

(12–20)

an

P

These two mutually perpendicular components are shown in Fig. 12–24f. Therefore, the magnitude of acceleration is the positive value of

a

at

Acceleration (f)

a = 2a2t + a2n

(12–21)

Fig. 12–24 (cont.)

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To better understand these results, consider the following two special cases of motion.

12 n

t

v an at

As the boy swings upward with a velocity v, his motion can be analyzed using n–t coordinates. As he rises, the magnitude of his velocity (speed) is decreasing, and so at will be negative. The rate at which the direction of his velocity changes is an, which is always positive, that is, towards the center of rotation.

1. If the particle moves along a straight line, then r S  and from # Eq. 12–20, an = 0. Thus a = at = v, and we can conclude that the tangential component of acceleration represents the time rate of change in the magnitude of the velocity. 2. If the particle moves along a curve with a constant speed, then # at = v = 0 and a = an = v2 >r. Therefore, the normal component of acceleration represents the time rate of change in the direction of the velocity. Since an always acts towards the center of curvature, this component is sometimes referred to as the centripetal (or center seeking) acceleration. As a result of these interpretations, a particle moving along the curved path in Fig. 12–25 will have accelerations directed as shown.

a  at Change in direction of velocity

at Increasing speed

an an

a

a at

Change in magnitude of velocity

Fig. 12–25

b

osculating plane

O

n s

O¿

ub un ut

Fig. 12–26

t

Three-Dimensional Motion. If the particle moves along a space curve, Fig. 12–26, then at a given instant the t axis is uniquely specified; however, an infinite number of straight lines can be constructed normal to the tangent axis. As in the case of planar motion, we will choose the positive n axis directed toward the path’s center of curvature O. This axis is referred to as the principal normal to the curve. With the n and t axes so defined, Eqs. 12–15 through 12–21 can be used to determine v and a. Since ut and un are always perpendicular to one another and lie in the osculating plane, for spatial motion a third unit vector, ub , defines the binormal axis b which is perpendicular to ut and un , Fig. 12–26. Since the three unit vectors are related to one another by the vector cross product, e.g., ub = ut * un , Fig. 12–26, it may be possible to use this relation to establish the direction of one of the axes, if the directions of the other two are known. For example, no motion occurs in the ub direction, and if this direction and ut are known, then un can be determined, where in this case un = ub * ut , Fig. 12–26. Remember, though, that un is always on the concave side of the curve.

12.7

Curvilinear motion: normal and tanGential Components

59

procedure for analysis

12

Coordinate System. • Provided the path of the particle is known, we can establish a set of n and t coordinates having a fixed origin, which is coincident with the particle at the instant considered. • The positive tangent axis acts in the direction of motion and the positive normal axis is directed toward the path’s center of curvature. Velocity. • The particle’s velocity is always tangent to the path. • The magnitude of velocity is found from the time derivative of the path function. # v = s tangential acceleration. • The tangential component of acceleration is the result of the time rate of change in the magnitude of velocity. This component acts in the positive s direction if the particle’s speed is increasing or in the opposite direction if the speed is decreasing. • The relations between at , v, t, and s are the same as for rectilinear motion, namely, # at = v at ds = v dv

Once the rotation is constant, the riders will then have only a normal component of acceleration.

Please refer to the Companion Website for the animation: The Dynamics of a Turning Vehicle Illustrating Force, f

• If at is constant, at = (at)c , the above equations, when integrated, yield

s = s0 + v0t + 12(at)ct2 v = v0 + (at)ct v2 = v20 + 2(at)c(s - s0) normal acceleration. • The normal component of acceleration is the result of the time rate of change in the direction of the velocity. This component is always directed toward the center of curvature of the path, i.e., along the positive n axis. • The magnitude of this component is determined from v2 an = r

• If the path is expressed as y = f(x), the radius of curvature r at any point on the path is determined from the equation r =

[1 + (dy>dx)2]3>2

 d2y>dx2  The derivation of this result is given in any standard calculus text.

Motorists traveling along this cloverleaf interchange experience a normal acceleration due to the change in direction of their velocity. A tangential component of acceleration occurs when the cars’ speed is increased or decreased.

60

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12.14

ExamplE

When the skier reaches point A along the parabolic path in Fig. 12–27a, he has a speed of 6 m>s which is increasing at 2 m>s2. Determine the direction of his velocity and the direction and magnitude of his acceleration at this instant. Neglect the size of the skier in the calculation.

Please refer to the Companion Website for the animation: The Forward Velocity of a Skateboarder at Different Heights

Solution Coordinate System. Although the path has been expressed in terms of its x and y coordinates, we can still establish the origin of the n, t axes at the fixed point A on the path and determine the components of v and a along these axes, Fig. 12–27a. Velocity. By definition, the velocity is always directed tangent to 1 2 1 the path. Since y = 20 x , dy>dx = 10 x, then at x = 10 m, dy>dx = 1. Hence, at A, v makes an angle of u = tan-11 = 45 with the x axis, Fig. 12–27b. Therefore, Ans. vA = 6 m>s 45 d # The acceleration is determined from a = vut + (v2 >r)un . However, it is first necessary to determine the radius of curvature of the path at A 1 (10 m, 5 m). Since d2y>dx2 = 10 , then

y  1 x2 20

y

n

r =

u vA

A

t

x

(a)

 d2y>dx2 

The acceleration becomes

5m

10 m

[1 + (dy>dx)2]3>2

=

31

+

1 101 x 2 2 4 3>2 1  10 

`

x = 10 m

= 28.28 m

v2 # aA = vut + u r n (6 m>s)2 = 2ut + u 28.28 m n = 5 2ut + 1.273un 6 m>s2

As shown in Fig. 12–27b, n

1.273 m/s2 90 45 f

a 2 m/s2

t

a = 2(2 m>s2)2 + (1.273 m>s2)2 = 2.37 m>s2 2 f = tan-1 = 57.5 1.273 Thus, 45 + 90 + 57.5 - 180 = 12.5 so that, a = 2.37 m>s2 12.5 d

(b)

Fig. 12–27

Ans.

NOTE: By using n, t coordinates, we were able to readily solve this problem through the use of Eq. 12–18, since it accounts for the separate changes in the magnitude and direction of v.

12.7

ExamplE

Curvilinear Motion: norMal and tangential CoMponents

12.15

12

A race car C travels around the horizontal circular track that has a radius of 300 m, Fig. 12–28. If the car increases its speed at a constant rate of 1.5 m>s2, starting from rest, determine the time needed for it to reach an acceleration of 2 m>s2. What is its speed at this instant? C an

at

n t

61

Please refer to the Companion Website for the animation: The Dynamics of a Turning Vehicle Illustrating Force, f

a r  300 m

Fig. 12–28

Solution Coordinate System. The origin of the n and t axes is coincident with the car at the instant considered. The t axis is in the direction of motion, and the positive n axis is directed toward the center of the circle. This coordinate system is selected since the path is known. Acceleration. The magnitude of acceleration can be related to its components using a = 2a2t + a2n . Here at = 1.5 m>s2. Since an = v2 >r, the velocity as a function of time must be determined first. v = v0 + (at)ct v = 0 + 1.5t Thus (1.5t)2 v2 an = = 0.0075t2 m>s2 = r 300 The time needed for the acceleration to reach 2 m>s2 is therefore a = 2a2t + a2n

2 m>s2 = 2(1.5 m>s2)2 + (0.0075t2)2

Solving for the positive value of t yields

0.0075t2 = 2(2 m>s2)2 - (1.5 m>s2)2 t = 13.28 s = 13.3 s

Ans.

Velocity. The speed at time t = 13.28 s is v = 1.5t = 1.5(13.28) = 19.9 m>s Ans. NOTE: Remember the velocity will always be tangent to the path, whereas the acceleration will be directed within the curvature of the path.

62

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12.16

ExamplE

The boxes in Fig. 12–29a travel along the industrial conveyor. If a box as in Fig. 12–29b starts from rest at A and increases its speed such that at = (0.2t) m>s2, where t is in seconds, determine the magnitude of its acceleration when it arrives at point B. Solution Coordinate System. The position of the box at any instant is defined from the fixed point A using the position or path coordinate s, Fig. 12–29b. The acceleration is to be determined at B, so the origin of the n, t axes is at this point. # Acceleration. To determine the acceleration components at = v # and an = v2 >r, it is first necessary to formulate v and v so that they may be evaluated at B. Since vA = 0 when t = 0, then # at = v = 0.2t (1)

(a) A s

L0

3m

v

dv =

L0

t

0.2t dt

v = 0.1t2

2m

n t

(2)

The time needed for the box to reach point B can be determined by realizing that the position of B is sB = 3 + 2p(2)>4 = 6.142 m, Fig. 12–29b, and since sA = 0 when t = 0 we have ds v = = 0.1t2 dt L0

B (b)

6.142 m

tB

0.1t2dt L0 6.142 m = 0.0333t3B ds =

tB = 5.690 s Substituting into Eqs. 1 and 2 yields # (aB)t = vB = 0.2(5.690) = 1.138 m>s2 n 5.242 m/s

2

vB = 0.1(5.69)2 = 3.238 m>s

aB

At B, rB = 2 m, so that t B

1.138 m/s2

(c)

Fig. 12–29

(aB)n =

(3.238 m>s)2 v2B = = 5.242 m>s2 rB 2m

The magnitude of aB , Fig. 12–29c, is therefore aB = 2(1.138 m>s2)2 + (5.242 m>s2)2 = 5.36 m>s2

Ans.

63

Curvilinear Motion: norMal and tangential CoMponents

12.7

PreliminAry Problem

12

P12–7.

d) Determine the normal and tangential components of acceleration at s = 0 if v = (4s + 1) m > s, where s is in meters.

a) Determine the acceleration at the instant shown. v  2 m/s v  3 m/s2

s

2m

1m

#

b) Determine the increase in speed and the normal component of acceleration at s = 2 m. At s = 0, v = 0.

e) Determine the acceleration at s = 2 m if v = (2 s) m > s2, where s is in meters. At s = 0, v = 1 m > s. s

s2m

v  4 m/s2

3m

2m

c) Determine the acceleration at the instant shown. The particle has a constant speed of 2 m > s.

f)

Determine the acceleration when t = 1 s if v = (4t2 + 2) m > s, where t is in seconds. v (4t2 + 2) m/s

y

y  2 x2 6m x 2 m/s

Prob. P12–7

64

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FundAmentAl ProblemS F12–27. The boat is traveling along the circular path with a speed of v = (0.0625t2) m>s, where t is in seconds. Determine the magnitude of its acceleration when t = 10 s. t

F12–30. When x = 3 m, the crate has a speed of 6 m>s which is increasing at 2 m>s2. Determine the direction of the crate’s velocity and the magnitude of the crate’s acceleration at this instant. y

v  0.0625t2 40 m

y  1 x2 8

n 6 m/s

O

Prob. F12–27 x

F12–28. The car is traveling along the road with a speed of v = (2s) m>s, where s is in meters. Determine the magnitude of its acceleration when s = 10 m.

3m

Prob. F12–30 v  (2s) m/s t

F12–31. If the motorcycle has a deceleration of at = -(0.001s) m>s2 and its speed at position A is 25 m>s, determine the magnitude of its acceleration when it passes point B. s

50 m n

A 90

s

O

300 m n

Prob. F12–28

B

F12–29. If the car decelerates uniformly along the curved road from 25 m>s at A to 15 m>s at C, determine the acceleration of the car at B.

t

Prob. F12–31 F12–32. The car travels up the hill with a speed of v = (0.2s) m>s, where s is in meters, measured from A. Determine the magnitude of its acceleration when it is at point s = 50 m, where r = 500 m.

A

250 m

y

rB  300 m

n

50 m

B C

A

s  50 m

t x

O

Prob. F12–29

Prob. F12–32

12.7

65

Curvilinear Motion: norMal and tangential CoMponents

ProblemS 12–110. The motion of a particle is defined by the equations x = (2t + t2) m and y = (t2) m, where t is in seconds. Determine the normal and tangential components of the particle’s velocity and acceleration when t = 2 s.

12–111. Determine the maximum constant speed a race car can have if the acceleration of the car cannot exceed 7.5 m>s2 while rounding a track having a radius of curvature of 200 m.

12 12–115. When the car reaches point A it has a speed of 25  m>s. If the brakes are applied, its speed is reduced by at = 1 - 14 t1>2 2 m>s2. Determine the magnitude of acceleration of the car just before it reaches point C.

*12–116. When the car reaches point A, it has a speed of 25 m>s. If the brakes are applied, its speed is reduced by at = (0.001s - 1) m>s2. Determine the magnitude of acceleration of the car just before it reaches point C.

250 m

r C

B

*12–112. A particle moves along the curve y = sin x with a constant speed v = 2 m>s. Determine the normal and tangential components of its velocity and acceleration at any instant.

A 200 m

30

Probs. 12–115/116

12–113. The position of a particle is defined by r = {4(t - sin t)i + (2t2 - 3)j} m, where t is in seconds and the argument for the sine is in radians. Determine the speed of the particle and its normal and tangential components of acceleration when t = 1 s.

12–114. The car travels along the curve having a radius of 300 m. If its speed is uniformly increased from 15 m>s to 27 m>s in 3 s, determine the magnitude of its acceleration at the instant its speed is 20 m>s.

12–117. At a given instant, a car travels along a circular curved road with a speed of 20 m>s while decreasing its speed at the rate of 3 m>s2. If the magnitude of the car’s acceleration is 5 m>s2, determine the radius of curvature of the road. 12–118. Car B turns such that its speed is increased by (at)B = (0.5et) m>s2 , where t is in seconds. If the car starts from rest when u = 0, determine the magnitudes of its velocity and acceleration when the arm AB rotates u = 30. Neglect the size of the car. v B

20 m/s

5m

300 m A

Prob. 12–114

u

Prob. 12–118

66

Chapter 12

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of a

partiCle

12–119. The motorcycle is traveling at 1 m>s when it is . 12 at A. If the speed is then increased at v = 0.1 m>s2, determine its speed and acceleration at the instant t = 5 s.

12–122. The car travels along the circular path such that its speed is increased by at = (0.5et) m>s2, where t is in seconds. Determine the magnitudes of its velocity and acceleration after the car has traveled s = 18 m starting from rest. Neglect the size of the car.

y y  0.5x2

s  18 m

s x

A

Prob. 12–119

ρ  30 m

Prob. 12–122 *12–120. The car passes point A with a speed of 25 m>s after which its speed is defined by v = (25 - 0.15s) m>s. Determine the magnitude of the car’s acceleration when it reaches point B, where s = 51.5 m and x = 50 m.

12–121. If the car passes point A with a speed of 20 m>s and begins to increase its speed at a constant rate of at = 0.5 m>s2, determine the magnitude of the car’s acceleration when s = 101.68 m and x = 0.

y y  16 

12–123. The satellite S travels around the earth in a circular path with a constant speed of 20 Mm>h. If the acceleration is 2.5 m>s2, determine the altitude h. Assume the earth’s diameter to be 12 713 km.

S

h

1 2 x 625

B

s A

Probs. 12–120/121

x

Prob. 12–123

12.7

67

Curvilinear Motion: norMal and tangential CoMponents

*12–124. The car has an initial speed v0 = 20 m>s. If it increases its speed along the circular track at s = 0, at = (0.8s) m>s2, where s is in meters, determine the time needed for the car to travel s = 25 m.

12–127. When the roller coaster is at B, it has a speed of 25  m>s, which is increasing at at = 3 m>s2. Determine the 12 magnitude of the acceleration of the roller coaster at this instant and the direction angle it makes with the x axis.

12–125. The car starts from rest at s = 0 and increases its speed at at = 4 m>s2. Determine the time when the magnitude of acceleration becomes 20 m>s2. At what position s does this occur?

*12–128. If the roller coaster starts from rest at A and its speed increases at at = (6 - 0.06s) m>s2, determine the magnitude of its acceleration when it reaches B where sB = 40 m.

y

y

s

1 x2 100

A s

B

r  40 m

30 m

Probs. 12–124/125 12–126. At a given instant the train engine at E has a speed of 20 m>s and an acceleration of 14 m>s2 acting in the direction shown. Determine the rate of increase in the train’s speed and the radius of curvature r of the path.

v  20 m/s

Probs. 12–127/128

12–129. The box of negligible size is sliding down along a curved path defined by the parabola y = 0.4x2. When it is at A(xA = 2 m, yA = 1.6 m), the speed is v = 8 m>s and the increase in speed is dv>dt = 4 m>s2. Determine the magnitude of the acceleration of the box at this instant.

75 2

a  14 m/s

E

y

r A 2

y  0.4x

x 2m

Prob. 12–126

Prob. 12–129

68

Chapter 12

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of a

partiCle

12–130. The position of a particle traveling along a curved 3 2 12 path is s = (3t - 4t + 4) m, where t is in seconds. When t = 2 s, the particle is at a position on the path where the radius of curvature is 25 m. Determine the magnitude of the particle’s acceleration at this instant.

12–135. Starting from rest, a bicyclist travels around a horizontal circular path, r = 10 m, at a speed of v = (0.09t2 + 0.1t) m>s, where t is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled s = 3 m.

12–131. A particle travels around a circular path having a radius of 50 m. If it is initially traveling with a speed of 10 m>s and its speed then increases at a rate of # v = (0.05 v) m>s2, determine the magnitude of the particle’s acceleration four seconds later.

*12–136. The motorcycle is traveling at a constant speed of 60 km>h. Determine the magnitude of its acceleration when it is at point A.

*12–132. The motorcycle is traveling at 40 m>s when it is # at A. If the speed is then decreased at v = - (0.05 s) m>s2, where s is in meters measured from A, determine its speed and acceleration when it reaches B.

y

y2  2x A

x 60

25 m 150 m B

150 m

Prob. 12–136 12–137. When t = 0, the train has a speed of 8 m>s, which is increasing at 0.5 m>s2. Determine the magnitude of the acceleration of the engine when it reaches point A, at t = 20 s. Here the radius of curvature of the tracks is rA = 400 m.

A

Prob. 12–132 12–133. At a given instant the jet plane has a speed of 550 m>s and an acceleration of 50 m>s2 acting in the direction shown. Determine the rate of increase in the plane’s speed, and also the radius of curvature r of the path.

vt  8 m/s A

550 m/s

Prob. 12–137 70 a  50

m/s2

r

12–138. The ball is ejected horizontally from the tube with a speed of 8 m>s. Find the equation of the path, y = f(x), and then find the ball’s velocity and the normal and tangential components of acceleration when t = 0.25 s. y vA  8 m/s

Prob. 12–133 12–134. A boat is traveling along a circular path having a radius of 20 m. Determine the magnitude of the boat’s acceleration when the speed is v = 5 m>s and the rate of # increase in the speed is v = 2 m>s2.

A

Prob. 12–138

x

Curvilinear Motion: norMal and tangential CoMponents

12.7

12–139. The motorcycle travels along the elliptical track at a constant speed v. Determine its greatest acceleration if a 7 b. *12–140. The motorcycle travels along the elliptical track at a constant speed v. Determine its smallest acceleration if a 7 b.

69

12–142. The ball is kicked with an initial speed vA = 8 m>s at an angle uA = 40 with the horizontal. Find the equation 12 of the path, y = f(x), and then determine the ball’s velocity and the normal and tangential components of its acceleration when t = 0.25 s.

y

y

b

vA = 8 m/s 2

y

a x

b

2

a

2

x

2

1

uA A

y

40

x

x

Prob. 12–142 Probs. 12–139/140

12–141. The race car has an initial speed vA = 15 m>s at A. If it increases its speed along the circular track at the rate at = (0.4s) m>s2, where s is in meters, determine the time needed for the car to travel 20 m. Take r = 150 m.

12–143. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the minimum acceleration experienced by the passengers.

*12–144. Cars move around the “traffic circle” which is in the shape of an ellipse. If the speed limit is posted at 60 km>h, determine the maximum acceleration experienced by the passengers.

y

r

2

x  y 1 (60)2 (40) 0)2 2

40 m s A

x 60 m

Prob. 12–141

Probs. 12–143/144

70

Chapter 12

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of a

partiCle

12–145. The particle travels with a constant speed of 12 300 mm>s along the curve. Determine the particle’s acceleration when it is located at point (200 mm, 100 mm) and sketch this vector on the curve.

*12–148. The jet plane is traveling with a speed of 120 m>s which is decreasing at 40 m>s2 when it reaches point A. Determine the magnitude of its acceleration when it is at this point. Also, specify the direction of flight, measured from the x axis. 12–149. The jet plane is traveling with a constant speed of 110 m>s along the curved path. Determine the magnitude of the acceleration of the plane at the instant it reaches point A(y = 0).

y (mm)

y

y

y  15 lnQ

20(103) x 80 m

x R 80

x

A v P x (mm)

Probs. 12–148/149

Prob. 12–145 12–146. The train passes point B with a speed of 20 m>s which is decreasing at at = - 0.5 m>s2. Determine the magnitude of acceleration of the train at this point. 12–147. The train passes point A with a speed of 30 m>s and begins to decrease its speed at a constant rate of at = - 0.25 m>s2. Determine the magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m.

y y  200

12–150. Particles A and B are traveling counterclockwise around a circular track at a constant speed of 8 m>s. If at the instant shown the speed of A begins to increase by (at)A = (0.4sA) m>s2, where sA is in meters, determine the distance measured counterclockwise along the track from B to A when t = 1 s. What is the magnitude of the acceleration of each particle at this instant? 12–151. Particles A and B are traveling around a circular track at a speed of 8 m>s at the instant shown. If the speed of B is increasing by (at)B = 4 m>s2, and at the same instant A has an increase in speed of (at)A = 0.8t m>s2, determine how long it takes for a collision to occur. What is the magnitude of the acceleration of each particle just before the collision occurs? A

x e 1000

sA u  120

B

A

B r5m x

400 m

Probs. 12–146/147

sB

Probs. 12–150/151

12.8 *12–152. A particle P moves along the curve y = (x2 - 4) m with a constant speed of 5 m>s. Determine the point on the curve where the maximum magnitude of acceleration occurs and compute its value.

12–153. When the bicycle passes point A, it has a speed of # 6 m>s, which is increasing at the rate of v = (0.5) m>s2. Determine the magnitude of its acceleration when it is at point A.

y

y

12 ln (

71

Curvilinear Motion: CylindriCal CoMponents

12–154. A particle P travels along an elliptical spiral path such that its position vector r is defined by 12 r = 5 2 cos(0.1t)i + 1.5 sin(0.1t)j + (2t)k 6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. When t = 8 s, determine the coordinate direction angles a, b, and g, which the binormal axis to the osculating plane makes with the x, y, and z axes. Hint: Solve for the velocity vP and acceleration aP of the particle in terms of their i, j, k components. The binormal is parallel to vP * aP. Why? z

x ) 20

P r

A

y x 50 m

x

Probs. 12–152/153

12.8

Prob. 12–154

Curvilinear Motion: Cylindrical Components

Sometimes the motion of the particle is constrained on a path that is best described using cylindrical coordinates. If motion is restricted to the plane, then polar coordinates are used.

u uu

Polar Coordinates.

We can specify the location of the particle shown in Fig. 12–30a using a radial coordinate r, which extends outward from the fixed origin O to the particle, and a transverse coordinate u, which is the counterclockwise angle between a fixed reference line and the r axis. The angle is generally measured in degrees or radians, where 1 rad = 180>p. The positive directions of the r and u coordinates are defined by the unit vectors ur and uu , respectively. Here ur is in the direction of increasing r when u is held fixed, and uu is in a direction of increasing u when r is held fixed. Note that these directions are perpendicular to one another.

r ur

r u O

Position (a)

Fig. 12–30

72

Chapter 12

K i n e M at i C s

At any instant the position of the particle, Fig. 12–30a, is defined by the position vector

uu

r ur

u Position

u¿r

u

(12–22)

# To evaluate ur , notice that ur only changes its direction with respect to time, since by definition the magnitude of this vector is always one unit. Hence, during the time t, a change r will not cause a change in the direction of ur ; however, a change u will cause ur to become ur= , where ur= = ur + ur , Fig. 12–30b. The time change in ur is then ur . For small angles u this vector has a magnitude ur  1(u) and acts in the uu direction. Therefore, ur = uuu , and so

(a)

uu

r = r ur

Velocity. The instantaneous velocity v is obtained by taking the time derivative of r. Using a dot to represent the time derivative, we have # # # v = r = r ur + r ur

r O

partiCle

Position.

u

12

of a

ur ur

(b)

ur u # ur = lim = a lim b uu t S 0 t t S 0 t # # ur = uuu

(12–23)

Substituting into the above equation, the velocity can be written in component form as v = vrur + vuuu

(12–24)

# vr = r # vu = ru

(12–25)

where

v vu vr r u O Velocity (c)

Fig. 12–30 (cont.)

These components are shown graphically in Fig. 12–30c. The radial component vr is a measure of the rate of increase or decrease in the # length of the radial coordinate, i.e., r ; whereas the transverse component vu can be interpreted as the rate of motion along #the circumference of a circle having a radius r. In particular, the term u = du>dt is called the angular velocity, since it indicates the time rate of change of the angle u. Common units used for this measurement are rad>s. Since vr and vu are mutually perpendicular, the magnitude of velocity or speed is simply the positive value of # # v = 2(r)2 + (ru)2

(12–26)

and the direction of v is, of course, tangent to the path, Fig. 12–30c.

12.8

73

Curvilinear Motion: CylindriCal CoMponents

Acceleration. Taking the time derivatives of Eq. 12–24, using Eqs. 12–25, we obtain the particle’s instantaneous acceleration,

12

$ ## $ # ## # # a = v = rur + rur + ruuu + ru uu + ruuu # To evaluate uu , it is necessary only to find the change in the direction of uu since its magnitude is always unity. During the time t, a change r will not change the direction of uu , however, a change u will cause uu to become uu= , where uu= = uu + uu , Fig. 12–30d. The time change in uu is thus uu . For small angles this vector has a magnitude uu  1(u) and acts in the -ur direction; i.e., uu = - uur . Thus, uu u # uu = lim = - a lim bur t S 0 t t S 0 t # # uu = -uur

uu

uu

u¿u ur u (d)

(12–27)

Substituting this result and Eq. 12–23 into the above equation for a, we can write the acceleration in component form as

a = a rur + a uuu

(12–28)

# $ ar = r - ru2 $ # # au = ru + 2ru

(12–29)

where

$ The term u = d2u>dt2 = d>dt(du>dt) is called the angular acceleration since it measures the change made in the angular velocity during an instant of time. Units for this measurement are rad>s2. Since ar and au are always perpendicular, the magnitude of acceleration is simply the positive value of # $ $ # # a = 2(r - r u 2)2 + (ru + 2r u)2

a au ar r u

(12–30)

The direction is determined from the vector addition of its two components. In general, a will not be tangent to the path, Fig. 12–30e.

O Acceleration (e)

74

Chapter 12

K i n e M at i C s uz

12

uu ur rP z O u

r

of a

partiCle

Cylindrical Coordinates. If the particle moves along a space curve as shown in Fig. 12–31, then its location may be specified by the three cylindrical coordinates, r, u, z. The z coordinate is identical to that used for rectangular coordinates. Since the unit vector defining its direction, uz , is constant, the time derivatives of this vector are zero, and therefore the position, velocity, and acceleration of the particle can be written in terms of its cylindrical coordinates as follows: rP = rur + zuz # # # v = rur + ruuu + zuz # $ $ # # $ a = (r - ru2)ur + (ru + 2ru)uu + zuz

(12–31) (12–32)

Time Derivatives. # $ # Fig. 12–31

The $ above equations require that we obtain the time derivatives r, r, u, and u in order to evaluate the r and u components of v and a. Two types of problems generally occur: 1. If the polar coordinates are specified as time parametric equations, r = r(t) and u = u(t), then the time derivatives can be found directly. 2. If the time-parametric equations are not given, then the path r = f(u) must be known. Using the# chain rule of calculus$we can then find the # $ relation between r and u, and between r and u . Application of the chain rule, along with some examples, is explained in Appendix C.

Procedure for Analysis Coordinate System.

• Polar coordinates are a suitable choice for solving problems when The spiral motion of this girl can be followed by using cylindrical components. Here the radial coordinate r is constant, the transverse coordinate u will increase with time as the  girl rotates about the vertical, and her altitude z will decrease with time.

data regarding the angular motion of the radial coordinate r is given to describe the particle’s motion. Also, some paths of motion can conveniently be described in terms of these coordinates.

• To use polar coordinates, the origin is established at a fixed point, and the radial line r is directed to the particle.

• The transverse coordinate u is measured from a fixed reference line to the radial line.

Velocity and acceleration.

# $ #

$

• Once r and the four time derivatives r, r, u, and u have been evaluated at the instant considered, their values can be substituted into Eqs. 12–25 and 12–29 to obtain the radial and transverse components of v and a.

• If it is necessary to take the time derivatives of r = f(u), then the chain rule of calculus must be used. See Appendix C.

a simple extension of the • Motion in three dimensions# requires $ above procedure to include z and z.

12.8

ExamplE

Curvilinear Motion: CylindriCal CoMponents

12.17

12

The amusement park ride shown in Fig. 12–32a consists of a chair that is rotating in a horizontal #circular path of radius r such$ that the arm OB has an angular velocity u and angular acceleration u . Determine the radial and transverse components of velocity and acceleration of the passenger. Neglect his size in the calculation.

n

O

· ·· u, u

r

· v  ru

· ar  ru 2

u B

r

·· au  ru

r

u, t

(a)

(b)

Fig. 12–32

Solution Coordinate System. Since the angular motion of the arm is reported, polar coordinates are chosen for the solution, Fig. 12–32a. Here u is not related to r, since the radius is constant for all u. Velocity and Acceleration. It is first necessary to specify the first and second time derivatives of r and u. Since r is constant, we have # $ r = r r = 0 r = 0 Thus,

75

# vr = r = 0 # vu = ru # # $ ar = r - ru2 = -ru2 $ $ # # au = ru + 2ru = ru

Ans. Ans. Ans. Ans.

These results are shown in Fig. 12–32b. NOTE: The n, t axes are also shown in Fig. 12–32b, which in this special

case of circular motion happen to be # collinear with the r and u axes, respectively. Since v = vu = vt = ru, then by comparison, # # (ru)2 v2 -ar = an = = = ru2 r r # $ dv d # dr # du au = at = = (ru) = u + r = 0 + ru dt dt dt dt

76

12

Chapter 12

ExamplE

K i n e M at i C s

of a

partiCle

12.18 The rod OA in Fig. 12–33a rotates in the horizontal plane such that u = (t3) rad. At the same time, the collar B is sliding outward along OA so that r = (100t2) mm. If in both cases t is in seconds, determine the velocity and acceleration of the collar when t = 1 s.

O

Solution Coordinate System. Since time-parametric equations of the path are given, it is not necessary to relate r to u.

r

B

u

Velocity and Acceleration. Determining the time derivatives and evaluating them when t = 1 s, we have A

r = 100t2 `

(a)

u  57.3

# r = 200t `

vu  300 mm/s

$ r = 200 `

u

t=1 s

t=1 s

t=1 s

= 100 mm u = t3 `

# = 200 mm>s u = 3t2 `

= 200 mm>s2

As shown in Fig. 12–33b, # # v = rur + ruuu

v d vr  200 mm/s

t=1 s

$ u = 6t `

= 1 rad = 57.3

t=1 s

t=1 s

= 3 rad>s = 6 rad>s2.

= 200ur + 100(3)uu = 5 200ur + 300uu 6 mm>s

r

The magnitude of v is

(b)

v = 2(200)2 + (300)2 = 361 mm>s d = tan-1 a

300 b = 56.3 d + 57.3 = 114 200

Ans. Ans.

As shown in Fig. 12–33c, # $ # # $ a = ( r - ru2)ur + (ru + 2ru)uu u  57.3

a f

= [200 - 100(3)2]ur + [100(6) + 2(200)3]uu

u au  1800 mm/s2

The magnitude of a is

ar  700 mm/s2 r (c)

Fig. 12–33

= 5 -700ur + 1800uu 6 mm>s2

f = tan-1 a

a = 2(-700)2 + (1800)2 = 1930 mm>s2

1800 b = 68.7 (180 - f) + 57.3 = 169 700

Ans. Ans.

NOTE: The velocity is tangent to the path; however, the acceleration is directed within the curvature of the path, as expected.

12.8

ExamplE

77

Curvilinear Motion: CylindriCal CoMponents

12.19

12

The searchlight in Fig. 12–34a casts a spot of light along the face of a wall that is located 100 m from the searchlight. Determine the magnitudes of the velocity and acceleration at which the spot appears to travel across the wall # at the instant u = 45. The searchlight rotates at a constant rate of u = 4 rad>s.

u

r ·

100 m

u  4 rad/s

Solution Coordinate System. Polar coordinates will be used to solve this problem since the angular rate of the searchlight is given. To find the necessary time derivatives it is first necessary to relate r to u. From Fig. 12–34a,

(a)

r

r = 100>cos u = 100 sec u Velocity and Acceleration. Using the chain rule of calculus, noting that d(sec u) = sec u tan u du, and d(tan u) = sec2 u du, we have # # r = 100(sec u tan u)u# # # # $ 2 r = 100(sec u tan u)u(tan $ u)u + 100 sec u(sec u)u(u) + 100 sec u tan u(u $ # ) # = 100 sec u tan2 u (u)2 + 100 sec3u (u)2 + 100(sec u tan u)u # $ Since u = 4 rad>s = constant, then u = 0, and the above equations, when u = 45, become

vr

v u r

As shown in Fig. 12–34c, # $ $ # # a = ( r - ru2)ur + (ru + 2ru)uu = [6788.2 - 141.4(4)2]ur + [141.4(0) + 2(565.7)4]uu = 5 4525.5ur + 4525.5uu 6 m>s2 a = 2a2r + a2u = 2(4525.5)2 + (4525.5)2 = 6400 m>s2

100 m

vu u (b)

r = 100 sec 45 = 141.4 # r = 400 sec 45 tan 45 = 565.7 $ r = 1600 (sec 45 tan2 45 + sec3 45) = 6788.2 As shown in Fig. 12–34b, # # v = rur + ruuu = 565.7ur + 141.4(4)uu = 5 565.7ur + 565.7uu 6 m>s v = 2v2r + v2u = 2(565.7)2 + (565.7)2 = 800 m>s

u

r ar

a u r

u

100 m

au

Ans.

Ans. $ NOTE: It is also possible to find a without having to calculate r (or ar). As shown in Fig. 12–34d, since au = 4525.5 m>s2, then by vector resolution, a = 4525.5>cos 45 = 6400 m>s2.

u

(c)

a u  45 ar

au  4525.5 m/s2

(d)

Fig. 12–34

78

12

Chapter 12

K i n e M at i C s

of a

partiCle

12.20

ExamplE

r  0.15 (1  cos u) m

u

r

Due to the rotation of the forked rod, the ball in Fig. 12–35a travels around the slotted path, a portion of which is in the shape of a cardioid, r = 0.15(1 - cos u) m, where u is in radians. If the ball’s velocity is v = 1.2 m>s and its acceleration is a = # 9 m>s2 at the instant u = 180, determine the angular velocity u and angular $ acceleration u of the fork. Solution Coordinate System. This path is most unusual, and mathematically it is best expressed using polar coordinates, as$ done here, rather than # rectangular coordinates. Also, since u and u must be determined, then r, u coordinates are an obvious choice.

· ·· u, u

(a)

Velocity and Acceleration. The time derivatives of r and u can be determined using the chain rule. r = 0.15(1 - cos u) # # r = 0.15(sin u)u # # $ $ r = 0.15(cos u) u(u) + 0.15(sin u)u Evaluating these results at u = 180, we have # # $ r = 0.3 m r = 0 r = -0.15u2 # Since v = 1.2 m>s, using Eq. 12–26 to determine u yields # # v = 2(r)2 + (ru)2

# 1.2 = 2(0)2 + (0.3u)2 # u = 4 rad>s

Ans.

$ In a similar manner, u can be found using Eq. 12–30. # $ $ # # a = 2(r - ru2)2 + (ru + 2ru)2

r v  1.2 m/s a  9 m/s2

$ 9 = 2[ -0.15(4)2 - 0.3(4)2]2 + [0.3u + 2(0)(4)]2 $ (9)2 = ( -7.2)2 + 0.09 u 2 $ u = 18 rad>s2

Ans.

u (b)

Fig. 12–35

Vectors a and v are shown in Fig. 12–35b. NOTE: At this location, the u and t (tangential) axes will coincide. The

+n (normal) axis is directed to the right, opposite to +r.

12.8

79

Curvilinear Motion: CylindriCal CoMponents

FundAmentAl ProblemS F12–33. The car# has a speed of 15 m>s. Determine the angular velocity u of the radial line OA at this instant.

12 F12–36. Peg P is driven by the forked link OA along the path described by r = e u, where r is in meters. When u = p4 rad, the link has an angular # $ velocity and angular acceleration of u = 2 rad>s and u = 4 rad>s2. Determine the radial and transverse components of the peg’s acceleration at this instant.

A

r  eu

A

r  400 m P

r u

Prob. F12–33

O

F12–34. The platform is rotating about the vertical axis such that at any instant its angular position is u = (4t3/2) rad, where t is in seconds. A ball rolls outward along the radial groove so that its position is r = (0.1t3) m, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the ball when t = 1.5 s.

u u, u

O

Prob. F12–36 F12–37. The collars are pin connected at B and are free to move along rod OA and the curved guide OC having the shape of a cardioid, r = [0.2(1 # + cos u)] m. At u = 30, the angular velocity of OA is u = 3 rad>s. Determine the magnitude of the velocity of the collars at this point. A

u, u

r  0.2(l + cos u) m B

r u r

Prob. F12–34

F12–35. Peg P is driven by the fork link OA along the curved path described by r = (2u) m. At the instant u = p>4 rad, the # angular velocity $ and angular acceleration of the link are u = 3 rad>s and u = 1 rad>s2. Determine the magnitude of the peg’s acceleration at this instant.

u

O

C

u  3 rad/s Prob. F12–37 F12–38. At the instant u = 45, the athlete is running with a constant speed of 2 m>s. Determine the angular velocity at which the camera must turn in order to follow the motion. r  (30 cosec u) m v

A

r

P u, u

30 m

r u

u A

O

Prob. F12–35

u

Prob. F12–38

80

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ProblemS 12–155. If a particle’s position is described by the polar coordinates r = 4(1 + sin t) m and u = (2e - t) rad, where t is in seconds and the argument for the sine is in radians, determine the radial and transverse components of the particle’s velocity and acceleration when t = 2 s. *12–156. A particle travels around a limaçon, defined by the equation r = b - a cos u, where a and b are constants. Determine the particle’s radial and transverse components of velocity and acceleration as a function of u and its time derivatives. 12–157. A particle moves along #a circular path of radius 300 mm. If its angular velocity is u = (2t2) rad>s, where t is in seconds, determine the magnitude of the particle’s acceleration when t = 2 s.

12–161. If a particle’s position is described by the polar coordinates r = (2 sin 2u) m and u = (4t) rad, and where t is in seconds, determine the radial and transverse components of its velocity and acceleration when t = 1 s. 12–162. If a particle moves along a path such that r = (eat) m and u = t, where t is in seconds, plot the path r = f(u), and determine the particle’s radial and transverse components of velocity and acceleration. 12–163. A # radar gun at O rotates with the angular $ velocity  of u = 0.1 rad>s and angular acceleration of u = 0.025 rad>s2, at the instant u = 45°, as it follows the motion of the car traveling along the circular road having a radius of r = 200  m. Determine the magnitudes of velocity and acceleration of the car at this instant.

12–158. For a short time a rocket travels up and to the right at a constant speed of 800 m>s along the parabolic path y = 600 - 35x2. Determine the radial and transverse components of velocity of the rocket at the instant u = 60°, where u is measured counterclockwise from the x axis.

r  200 m

12–159. The box slides down the helical ramp with a constant speed of v = 2 m>s. Determine the magnitude of its acceleration. The ramp descends a vertical distance of 1 m for every full revolution.The mean radius of the ramp is r = 0.5 m. *12–160. The box slides down the helical ramp such that r = 0.5 m, u = (0.5t3) rad, and z = (2 - 0.2t2) m, where t is in seconds. Determine the magnitudes of the velocity and acceleration of the box at the instant u = 2p rad.

u O

Prob. 12–163 *12–164. The small washer is sliding down the cord OA. When it is at the midpoint, its speed is 28 m>s and its acceleration is 7 m>s2. Express the velocity and acceleration of the washer at this point in terms of its cylindrical components. z A

0.5 m

6m O

x

Probs. 12–159/160

3m

Prob. 12–164

2m

y

12.8 12–165. The time rate of change of acceleration is referred to as the jerk, which is often used as a means of measuring # passenger discomfort. Calculate this vector, a, in terms of its cylindrical components, using Eq. 12–32. 12–166. A particle is moving along a circular path having a 400-mm radius. Its position as a function of time is given by u = (2t2) rad, where t is in seconds. Determine the magnitude of the particle’s acceleration when u = 30°. The particle starts from rest when u = 0°.

Curvilinear Motion: CylindriCal CoMponents

12–169. At the instant shown, the man# is twirling a hose over his head with an$ angular velocity u = 2 rad>s and an 12 angular acceleration u = 3 rad>s2. If it is assumed that the hose lies in a horizontal plane, and water is flowing through it at a constant rate of 3 m>s, determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r = 1.5 m.

· u  2 rad/s ·· u  3 rad/s2

12–167. The slotted link is pinned at O, and as a result of # the constant angular velocity u = 3 rad>s it drives the peg P for a short distance along the spiral guide r = (0.4 u) m, where u is in radians. Determine the radial and transverse components of the velocity and acceleration of P at the instant u = p>3 rad.

0.5 m

81

r  1.5 m u

P

Prob. 12–169

r r  0.4u

· u  3 rad/s u O

Prob. 12–167

12–170. The partial surface of the cam is that of a logarithmic spiral r = (40e0.05u) mm, where u is in# radians. If the cam is rotating at a constant angular rate of u = 4 rad/s, determine the magnitudes of the velocity and acceleration of the follower rod at the instant u = 30°. 12–171. Solve Prob. 12–170, if the cam has an angular $ = 2 rad>s2 when its angular velocity is acceleration of u # u = 4 rad>s at u = 30°.

*12–168. At the instant shown, the watersprinkler is # rotating with $an angular speed u = 2 rad>s and an angular acceleration u = 3 rad>s2. If the nozzle lies in the vertical plane and water is flowing through it at a constant rate of 3 m>s, determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r = 0.2 m. u

r

r

0.2 m

u



Prob. 12–168

4 rad/s

Probs. 12–170/171

40e0.05u

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*12–172. A cameraman standing at A is following the 12 movement of a race car, B, which is traveling around a curved track# at a constant speed of 30 m>s. Determine the angular rate u at which the man must turn in order to keep the camera directed on the car at the instant u = 30°.

vB

12–175. A block moves outward along the slot in the # platform with a speed of r = (4t) m>s, where t is in seconds. The platform rotates at a constant rate of 6 rad>s. If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when t = 1 s.

·

u  6 rad/s r

30 m/s u

BB r r

2020 mm AA

2020 mm

Prob. 12–175

uu 2020 mm

2020 mm

*12–176. The pin follows the path described by the equation r# = (0.2 + 0.15 cos $ u) m. At the instant u = 30, u = 0.7 rad/s and u = 0.5 rad/s2. Determine the magnitudes of the pin’s velocity and acceleration at this instant. Neglect the size of the pin. y

Prob. 12–172

r

0.2

0.15 cos u

r

12–173. The car travels around the circular track with a constant speed of 20 m>s. Determine the car’s radial and transverse components of velocity and acceleration at the instant u = p>4 rad. 12–174. The car travels around the circular track such that its transverse component is u = (0.006t2) rad, where t is in seconds. Determine the car’s radial and transverse components of velocity and acceleration at the instant t = 4 s.

r  (400 cos u) m r

u

u

0.7 rad/s

Prob. 12–176 12–177. The rod OA rotates clockwise with a constant angular velocity of 6 rad>s. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 200(2 − cos u) mm. Determine the speed of the slider blocks at the instant u = 150°. 12–178. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–177 when u = 150°. A B r

u

x

6 rad/s u O

400 mm

600 mm 200 mm

Probs. 12–173/174

Probs. 12–177/178

12.8

Curvilinear Motion: CylindriCal CoMponents

83

12–179. The rod OA rotates counterclockwise with a # constant angular velocity of u = 5 rad>s. Two pin-connected slider blocks, located at B, move freely on OA and the curved rod whose shape is a limaçon described by the equation r = 100(2 − cos u) mm. Determine the speed of the slider blocks at the instant u = 120°.

12–183. If the cam # rotates clockwise with a constant angular velocity of u = 5 rad>s, determine the magnitudes of 12 the velocity and acceleration of the follower rod AB at the instant u = 30°. The surface of the cam has a shape of limaçon defined by r = (200 + 100 cos u) mm.

*12–180. Determine the magnitude of the acceleration of the slider blocks in Prob. 12–179 when u = 120°.

*12–184. At the instant u = 30°, # the cam rotates with a clockwise angular velocity of u = 5 rad>s and, angular $ acceleration of u = 6 rad/s2. Determine the magnitudes of the velocity and acceleration of the follower rod AB at this instant. The surface of the cam has a shape of a limaçon defined by r = (200 + 100 cos u) mm.

· u  5 rad/s

y

A B r

(200

r

u

100 cos u) mm

x

O

u

r  100 (2  cos u) mm

A

B

Probs. 12–183/184

Probs. 12–179/180 12–181. The slotted arm AB drives pin C through the spiral groove described by the# equation r = au. If the angular velocity is constant at u, determine the radial and transverse components of velocity and acceleration of the pin. 12–182. The slotted arm AB drives pin C through the spiral groove described by the equation r = (1.5 u) m, where u is in radians. If the arm starts from rest when u = 60° and is driven # at an angular velocity of u = (4t) rad>s, where t is in seconds, determine the radial and transverse components of velocity and acceleration of the pin C when t = 1 s.

12–185. A truck is traveling along the horizontal circular curve of radius r = 60 m with a constant# speed v = 20 m>s. Determine the angular rate of rotation u of the radial line r and the magnitude of the truck’s acceleration. 12–186. A truck is traveling along the horizontal circular curve of radius r = 60 m with a speed of 20 m>s which is increasing at 3 m>s2. Determine the truck’s radial and transverse components of acceleration.

B C

u

r

r  60 m u u

A

Probs. 12–181/182

Probs. 12–185/186

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12–187. The double collar C is pin connected together such 12 that one collar slides over the fixed rod and the other slides over the #rotating rod AB. If the angular velocity of AB is 2 given as u = (e0.5 t ) rad>s, where t is in seconds, and the path defined by the fixed rod is r = |(0.4 sin u + 0.2)| m, determine the radial and transverse components of the collar’s velocity and acceleration when t = 1 s. When t = 0, u = 0. Use Simpson’s rule with n = 50 to determine u at t = 1 s. *12–188. The double collar C is pin connected together such that one collar slides over the fixed rod and the other slides over the rotating rod AB. If the mechanism is to be designed so that the largest speed given to the collar # is 6 m>s, determine the required constant angular velocity u of rod AB. The path defined by the fixed rod is r = (0.4 sin u + 0.2) m.

12–191. The driver of the car maintains a constant speed of 40 m>s. Determine the angular velocity of the camera tracking the car when u = 15°. *12–192. When u = 15°, the car has a speed of 50 m>s which is increasing at 6 m>s2. Determine the angular velocity of the camera tracking the car at this instant.

r

(100 cos 2u) m

B u 0.6 m

C

r u

A 0.2 m

Probs. 12–191/192

0.2 m 0.2 m

Probs. 12–187/188 12–189. For a short distance the train travels along a track having the shape of a spiral, r = (1000/u) m, where u is in radians. If it maintains a constant speed v = 20 m>s, determine the radial and transverse components of its velocity when u = (9p/4) rad. 12–190. For a short distance the train travels along a track having the shape of a spiral, r = (1000/u) m,# where u is in radians. If the angular rate is constant, u = 0.2 rad>s, determine the radial and transverse components of its velocity and acceleration when u = (9p/4) rad.

12–193. If the circular plate# rotates clockwise with a constant angular velocity of u = 1.5 rad/s, determine the magnitudes of the velocity and acceleration of the follower rod AB when u = 2/3p rad. and 12–194. When u = 2/3p rad, the angular velocity # angular of the circular plate are u = 1.5 rad/s $ acceleration and u = 3 rad/s2, respectively. Determine the magnitudes of the velocity and acceleration of the rod AB at this instant.

u

r u A

B

r

r

1000 u r

Probs. 12–189/190

(10

1/2

50 u

) mm

Probs. 12–193/194

12.9

12.9

85

absolute dependent Motion analysis of two partiCles

Absolute Dependent Motion Analysis of Two Particles

12

In some types of problems the motion of one particle will depend on the corresponding motion of another particle. This dependency commonly occurs if the particles, here represented by blocks, are interconnected by inextensible cords which are wrapped around pulleys. For example, the movement of block A downward along the inclined plane in Fig. 12–36 will cause a corresponding movement of block B up the other incline. We can show this mathematically by first specifying the location of the blocks using position coordinates sA and sB . Note that each of the coordinate axes is (1) measured from a fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of each block, and (3) has a positive sense from the fixed datums to A and to B. If the total cord length is lT , the two position coordinates are related by the equation

Datum sA

C D

Datum sB

O A

B

Fig. 12–36

sA + lCD + sB = lT Here lCD is the length of the cord passing over arc CD. Taking the time derivative of this expression, realizing that lCD and lT remain constant, while sA and sB measure the segments of the cord that change in length, we have dsA dsB + = 0 dt dt

or

vB = -vA

The negative sign indicates that when block A has a velocity downward, i.e., in the direction of positive sA , it causes a corresponding upward velocity of block B; i.e., B moves in the negative sB direction. In a similar manner, time differentiation of the velocities yields the relation between the accelerations, i.e., aB = -aA A more complicated example is shown in Fig. 12–37a. In this case, the position of block A is specified by sA , and the position of the end of the cord from which block B is suspended is defined by sB . As above, we have chosen position coordinates which (1) have their origin at fixed points or datums, (2) are measured in the direction of motion of each block, and (3) from the fixed datums are positive to the right for sA and positive downward for sB. During the motion, the length of the red colored segments of the cord in Fig. 12–37a remains constant. If l represents the total length of cord minus these segments, then the position coordinates can be related by the equation 2sB + h + sA = l

Datum sB

B

h

A

Since l and h are constant during the motion, the two time derivatives yield 2vB = -vA

2aB = -aA

Hence, when B moves downward (+sB), A moves to the left (-sA) with twice the motion.

Datum

sA

(a)

Fig. 12–37

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This example can also be worked by defining the position of block B from the center of the bottom pulley (a fixed point), Fig. 12–37b. In this case

12 Datum

2(h - sB) + h + sA = l Time differentiation yields sB

2vB = vA B

h

Datum

Here the signs are the same. Why?

A

Datum

2aB = aA

sA

(b)

Fig. 12–37 (cont.)

Procedure for Analysis The above method of relating the dependent motion of one particle to that of another can be performed using algebraic scalars or position coordinates provided each particle moves along a rectilinear path. When this is the case, only the magnitudes of the velocity and acceleration of the particles will change, not their line of direction. position-Coordinate Equation.

• Establish each position coordinate with an origin located at a fixed point or datum.

• It is not necessary that the origin be the same for each of the

coordinates; however, it is important that each coordinate axis selected be directed along the path of motion of the particle.

• Using geometry or trigonometry, relate the position coordinates

to the total length of the cord, lT , or to that portion of cord, l, which excludes the segments that do not change length as the particles move—such as arc segments wrapped over pulleys.

• If a problem involves a system of two or more cords wrapped

around pulleys, then the position of a point on one cord must be related to the position of a point on another cord using the above procedure. Separate equations are written for a fixed length of each cord of the system and the positions of the two particles are then related by these equations (see Examples 12.22 and 12.23).

time Derivatives.

• Two successive time derivatives of the position-coordinate

equations yield the required velocity and acceleration equations which relate the motions of the particles.

• The signs of the terms in these equations will be consistent with The cable is wrapped around the pulleys on this crane in order to reduce the required force needed to hoist a load.

those that specify the positive and negative sense of the position coordinates.

12.9

ExamplE

absolute dependent Motion analysis of two partiCles

12.21

87

12

Determine the speed of block A in Fig. 12–38 if block B has an upward speed of 6 m>s.

C

D

Datum

sB

sA

E

B

6 m/s

A

Fig. 12–38

Solution Position-Coordinate equation. There is one cord in this system having segments which change length. Position coordinates sA and sB will be used since each is measured from a fixed point (C or D) and extends along each block’s path of motion. In particular, sB is directed to point E since motion of B and E is the same. The red colored segments of the cord in Fig. 12–38 remain at a constant length and do not have to be considered as the blocks move. The remaining length of cord, l, is also constant and is related to the changing position coordinates sA and sB by the equation sA + 3sB = l time derivative. Taking the time derivative yields vA + 3vB = 0 so that when vB = -6 m>s (upward), vA = 18 m>s T

Ans.

88

12

Chapter 12

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12.22 Determine the speed of A in Fig. 12–39 if B has an upward speed of 6 m>s. Datum sA sC A

sB C

D

6 m/s

B

Fig. 12–39

Solution Position-Coordinate equation. As shown, the positions of blocks A and B are defined using coordinates sA and sB . Since the system has two cords with segments that change length, it will be necessary to use a third coordinate, sC , in order to relate sA to sB . In other words, the length of one of the cords can be expressed in terms of sA and sC , and the length of the other cord can be expressed in terms of sB and sC . The red colored segments of the cords in Fig. 12–39 do not have to be considered in the analysis. Why? For the remaining cord lengths, say l1 and l2 , we have sA + 2sC = l1

sB + (sB - sC) = l2

time derivative. Taking the time derivative of these equations yields vA + 2vC = 0

2vB - vC = 0

Eliminating vC produces the relationship between the motions of each cylinder. vA + 4vB = 0 so that when vB = -6 m>s (upward), vA = +24 m>s = 24 m>s T

Ans.

12.9

ExamplE

absolute dependent Motion analysis of two partiCles

12.23

89

12

Determine the speed of block B in Fig. 12–40 if the end of the cord at A is pulled down with a speed of 2 m>s.

D Datum sC

C sB

sA

A E B

2 m/s

Fig. 12–40

Solution Position-Coordinate equation. The position of point A is defined by sA , and the position of block B is specified by sB since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fixed pin at pulley D. Since the system consists of two cords, the coordinates sA and sB cannot be related directly. Instead, by establishing a third position coordinate, sC , we can now express the length of one of the cords in terms of sB and sC , and the length of the other cord in terms of sA , sB , and sC . Excluding the red colored segments of the cords in Fig. 12–40, the remaining constant cord lengths l1 and l2 (along with the hook and link dimensions) can be expressed as sC + sB = l1 (sA - sC) + (sB - sC) + sB = l2 time derivative. The time derivative of each equation gives vC + vB = 0 vA - 2vC + 2vB = 0 Eliminating vC, we obtain vA + 4vB = 0 so that when vA = 2 m>s (downward), vB = -0.5 m>s = 0.5 m>s c

Ans.

90

12

Chapter 12

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12.24 A man at A is hoisting a safe S as shown in Fig. 12–41 by walking to the right with a constant velocity vA = 0.5 m>s. Determine the velocity and acceleration of the safe when it reaches the elevation of 10 m. The rope is 30 m long and passes over a small pulley at D.

D E 15 m 10 m

C S

y A x

Fig. 12–41

Solution Position-Coordinate equation. This problem is unlike the previous examples since rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the positions of C and A, are specified by means of the x and y coordinates since they must be measured from a fixed point and directed along the paths of motion of the ends of the rope. The x and y coordinates may be related since the rope has a fixed length l = 30 m, which at all times is equal to the length of segment DA plus CD. Using the Pythagorean theorem to determine lDA , we have lDA = 2(15)2 + x2; also, lCD = 15 - y. Hence, l = lDA + lCD 30 = 2(15)2 + x2 + (15 - y) vA  0.5 m/s y = 2225 + x2 - 15 (1) time derivatives. Taking the time derivative, using the chain rule (see Appendix C), where vS = dy>dt and vA = dx>dt, yields

dy 1 2x dx = J R 2 dt 2 2225 + x dt x = vA (2) 2225 + x2 At y = 10 m, x is determined from Eq. 1, i.e., x = 20 m. Hence, from Eq. 2 with vA = 0.5 m>s, 20 vS = (0.5) = 0.4 m>s = 400 mm>s c Ans. 2225 + (20)2 The acceleration is determined by taking the time derivative of Eq. 2. Since vA is constant, then aA = dvA >dt = 0, and we have -x(dx>dt) d2y dvA 225v2A 1 dx 1 aS = 2 = c d xv + c d a b v + c d x = A A dt dt (225 + x2)3>2 (225 + x2)3>2 2225 + x2 dt 2225 + x2 vS =

At x = 20 m, with vA = 0.5 m>s, the acceleration becomes aS =

225(0.5 m>s)2 [225 + (20 m)2]3>2

= 0.00360 m>s2 = 3.60 mm>s2 c

Ans.

NOTE: The constant velocity at A causes the other end C of the rope

to have an acceleration since vA causes segment DA to change its direction as well as its length.

12.10

91

relative-Motion of two partiCles using translating axes

12.10 Relative-Motion of Two Particles

12

Using Translating Axes

Throughout this chapter the absolute motion of a particle has been determined using a single fixed reference frame. There are many cases, however, where the path of motion for a particle is complicated, so that it may be easier to analyze the motion in parts by using two or more frames of reference. For example, the motion of a particle located at the tip of an airplane propeller, while the plane is in flight, is more easily described if one observes first the motion of the airplane from a fixed reference and then superimposes (vectorially) the circular motion of the particle measured from a reference attached to the airplane. In this section translating frames of reference will be considered for the analysis.

Position. Consider particles A and B, which move along the

arbitrary paths shown in Fig. 12–42. The absolute position of each particle, rA and rB , is measured from the common origin O of the fixed x, y, z reference frame. The origin of a second frame of reference x, y, z is attached to and moves with particle A. The axes of this frame are only permitted to translate relative to the fixed frame. The position of B measured relative to A is denoted by the relative-position vector rB>A . Using vector addition, the three vectors shown in Fig. 12–42 can be related by the equation rB = rA + rB>A

z¿

z

A

Translating observer rB/A

rA

Fixed observer

B y

O x¿

(12–33) x

Fig. 12–42

Velocity.

An equation that relates the velocities of the particles is determined by taking the time derivative of the above equation; i.e., vB = vA + vB>A

(12–34)

Here vB = drB >dt and vA = drA >dt refer to absolute velocities, since they are observed from the fixed frame; whereas the relative velocity vB>A = drB>A >dt is observed from the translating frame. It is important to note that since the x, y, z axes translate, the components of rB>A will not change direction and therefore the time derivative of these components will only have to account for the change in their magnitudes. Equation 12–34 therefore states that the velocity of B is equal to the velocity of A plus (vectorially) the velocity of “B with respect to A,” as measured by the translating observer fixed in the x, y, z reference frame.

y¿

rB

92

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Acceleration. The time derivative of Eq. 12–34 yields a similar vector relation between the absolute and relative accelerations of particles A and B. aB = aA + aB>A

(12–35)

Here aB>A is the acceleration of B as seen by the observer located at A and translating with the x, y, z reference frame.*

Procedure for Analysis • When applying the relative velocity and acceleration equations, it is first necessary to specify the particle A that is the origin for the translating x, y, z axes. Usually this point has a known velocity or acceleration.

• Since vector addition forms a triangle, there can be at most two unknowns, represented by the magnitudes and > or directions of the vector quantities.

• These unknowns can be solved for either graphically, using

trigonometry (law of sines, law of cosines), or by resolving each of the three vectors into rectangular or Cartesian components, thereby generating a set of scalar equations.

The pilots of these close-flying planes must be aware of their relative positions and velocities at all times in order to avoid a collision. * An easy way to remember the setup of these equations is to note the “cancellation” of the subscript A between the two terms, e.g., aB = aA + aB>A .

12.10

ExamplE

93

relative-Motion of two partiCles using translating axes

12.25

12

A train travels at a constant speed of 60 km>h and crosses over a road as shown in Fig. 12–43a. If the automobile A is traveling at 45 km>h along the road, determine the magnitude and direction of the velocity of the train relative to the automobile. Solution i Vector Analysis. The relative velocity vT>A is measured from the translating x, y axes attached to the automobile, Fig. 12–43a. It is determined from vT = vA + vT>A . Since vT and vA are known in both magnitude and direction, the unknowns become the x and y components of vT>A . Using the x, y axes in Fig. 12–43a, we have

45

T vT  60 km/h y¿

y x

A

x¿ vA  45 km/h (a)

vT = vA + vT>A 60i = (45 cos 45i + 45 sin 45j) + vT>A vT>A = 5 28.2i - 31.8j 6 km>h

The magnitude of vT>A is thus

vT>A = 2(28.2)2 + ( -31.8)2 = 42.5 km>h

Ans.

28.2 km/h

From the direction of each component, Fig. 12–43b, the direction of vT>A is (vT>A)y 31.8 tan u = = (vT>A)x 28.2 u = 48.5 Ans. Note that the vector addition shown in Fig. 12–43b indicates the correct sense for vT>A . This figure anticipates the answer and can be used to check it.

u

vT/A

31.8 km/h (b)

Solution ii Scalar Analysis. The unknown components of vT>A can also be determined by applying a scalar analysis. We will assume these components act in the positive x and y directions. Thus, vT = vA + vT>A c

60 km>h 45 km>h (v ) (v ) d = c d + c T>A x d + c T>A y d S S c a 45

Resolving each vector into its x and y components yields + ) (S 60 = 45 cos 45 + (vT>A)x + 0 (+ c ) 0 = 45 sin 45 + 0 + (vT>A)y Solving, we obtain the previous results, (vT>A)x = 28.2 km>h = 28.2 km>h S (vT>A)y = -31.8 km>h = 31.8 km>h T

vT/A

vA  45 km/h

45

u vT  60 km/h

(c)

Fig. 12–43

94

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ExamplE

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partiCle

12.26

y¿

y

700 km/h

600 km/h A

B

x¿

50 km/h2

of a

x

100 km/h2 400 km

4 km (a)

Plane A in Fig. 12–44a is flying along a straight-line path, whereas plane B is flying along a circular path having a radius of curvature of rB = 400 km. Determine the velocity and acceleration of B as measured by the pilot of A. Solution Velocity. The origin of the x and y axes are located at an arbitrary fixed point. Since the motion relative to plane A is to be determined, the translating frame of reference x, y is attached to it, Fig. 12–44a. Applying the relative-velocity equation in scalar form since the velocity vectors of both planes are parallel at the instant shown, we have (+ c )

vB = vA + vB>A 600 km>h = 700 km>h + vB>A vB>A = -100 km>h = 100 km>h T

vB/A vA  700 km/h v  600 km/h B

Ans.

The vector addition is shown in Fig. 12–44b. Acceleration. Plane B has both tangential and normal components of acceleration since it is flying along a curved path. From Eq. 12–20, the magnitude of the normal component is

(b)

(600 km>h)2 v2B = = 900 km>h2 r 400 km Applying the relative-acceleration equation gives (aB)n =

aB = aA + aB>A 900i - 100j = 50j + aB>A Thus, aB>A = 5 900i - 150j 6 km>h2

From Fig. 12–44c, the magnitude and direction of aB>A are therefore aB>A = 912 km>h2 u = tan-1 900 km/h2 u aB/A

150 km/h2 (c)

Fig. 12–44

150 = 9.46 900

d

Ans.

NOTE: The solution to this problem was possible using a translating frame of reference, since the pilot in plane A is “translating.” Observation of the motion of plane A with respect to the pilot of plane B, however, must be obtained using a rotating set of axes attached to plane B. (This assumes, of course, that the pilot of B is fixed in the rotating frame, so he does not turn his eyes to follow the motion of A.) The analysis for this case is given in Example 16.21.

12.10

ExamplE

12.27

12

At the instant shown in Fig. 12–45a, cars A and B are traveling with speeds of 18 m>s and 12 m>s, respectively. Also at this instant, A has a decrease in speed of 2 m>s2, and B has an increase in speed of 3 m>s2. Determine the velocity and acceleration of B with respect to A. Solution Velocity. The fixed x, y axes are established at an arbitrary point on the ground and the translating x, y axes are attached to car A, Fig.  12–45a. Why? The relative velocity is determined from vB = vA + vB>A . What are the two unknowns? Using a Cartesian vector analysis, we have

y¿

3 m/s2

18 m/s

r  100 m 12 m/s B y 60

-12j = (-18 cos 60i - 18 sin 60j) + vB>A vB>A = 5 9i + 3.588j 6 m>s

2 m/s2 60 x¿

A

vB = vA + vB>A

Thus,

95

relative-Motion of two partiCles using translating axes

x

(a)

vB>A = 2(9)2 + (3.588)2 = 9.69 m>s

Ans.

Noting that vB>A has +i and +j components, Fig. 12–45b, its direction is (vB>A)y

= (vB>A)x u = 21.7

3.588 9 d

tan u =

Ans.

3.588 m/s

Acceleration. Car B has both tangential and normal components of acceleration. Why? The magnitude of the normal component is v2B

u

2

9 m/s

(12 m>s) = 1.440 m>s2 r 100 m Applying the equation for relative acceleration yields (aB)n =

=

(b)

aB = aA + aB>A ( -1.440i - 3j) = (2 cos 60i + 2 sin 60j) + aB>A aB>A = 5 -2.440i - 4.732j 6 m>s2

Here aB>A has -i and -j components. Thus, from Fig. 12–45c, aB>A = 2(2.440)2 + (4.732)2 = 5.32 m>s2

tan f =

vB/A

(aB>A)y (aB>A)x

=

2.440 m/s2 f

Ans.

4.732 2.440

f = 62.7 d

Ans.

NOTE: Is it possible to obtain the relative acceleration of aA>B using this

method? Refer to the comment made at the end of Example 12.26.

aB/A

4.732 m/s2 (c)

Fig. 12–45

96

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FundAmentAl ProblemS F12–39. Determine the velocity of block D if end A of the rope is pulled down with a speed of vA = 3 m>s. B

F12–42. Determine the velocity of block A if end F of the rope is pulled down with a speed of vF = 3 m>s. C

C

E

B A

D

A

vA  3 m/s F D

vF  3 m/s

Prob. F12–39 F12–40. Determine the velocity of block A if end B of the rope is pulled down with a speed of 6 m>s.

Prob. F12–42

F12–43. Determine the velocity of car A if point P on the cable has a speed of 4 m>s when the motor M winds the cable in.

M 6 m/s

P

B A

A

Prob. F12–43 Prob. F12–40 F12–41. Determine the velocity of block A if end B of the rope is pulled down with a speed of 1.5 m>s.

F12–44. Determine the velocity of cylinder B if cylinder A moves downward with a speed of vA = 4 m>s. F

E

C

1.5 m/s B

D A vA

A

Prob. F12–41

B

4 m/s

Prob. F12–44

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relative-Motion of two partiCles using translating axes

F12–45. Car A is traveling with a constant speed of 80 km>h due north, while car B is traveling with a constant speed of 100 km>h due east. Determine the velocity of car B relative to car A.

F12–47. The boats A and B travel with constant speeds of vA = 15 m>s and vB = 10 m>s when they leave the pier at 12 O at the same time. Determine the distance between them when t = 4 s.

y vB  10 m/s B

B 100 km/h 2 km

A

30 O 45

vA  15 m/s

30 x

A 80 km/h

Prob. F12–47

Prob. F12–45

F12–46. Two planes A and B are traveling with the constant velocities shown. Determine the magnitude and direction of the velocity of plane B relative to plane A.

F12–48. At the instant shown, cars A and B are traveling at the speeds shown. If B is accelerating at 1200 km>h2 while A maintains a constant speed, determine the velocity and acceleration of A with respect to B.

45

vB  800 km/h B vA  650 km/h

A

60 20 km/h B

A

65 km/h

Prob. F12–46

Prob. F12–48

100 m

98

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ProblemS 12–195. If the end of the cable at A is pulled down with a speed of 5 m>s, determine the speed at which block B rises.

12–198. Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5 s. 12–199. Starting from rest, the cable can be wound onto the drum of the motor at a rate of vA = (3t2) m>s, where t is in seconds. Determine the time needed to lift the load 7 m.

A 5 m/s

A

B

D C B

Prob. 12–195 *12–196. Determine the displacement of the log if the truck at C pulls the cable 1.2 m to the right. Probs. 12–198/199 C

B

*12–200. If the end of the cable at A is pulled down with a speed of 2 m>s, determine the speed at which block B rises. D

C

A 2 m/s

Prob. 12–196 12–197. If the end A of the cable is moving upwards at vA = 14 m>s, determine the speed of block B.

B

F

Prob. 12–200

vA

14 m/s

A

E

D C

12–201. The motor at C pulls in the cable with an acceleration aC = (3t2) m>s2, where t is in seconds. The motor at D draws in its cable at aD = 5 m>s2. If both motors start at the same instant from rest when d = 3 m, determine (a) the time needed for d = 0, and (b) the velocities of blocks A and B when this occurs. D B A

B d

Prob. 12–197

Prob. 12–201

C

12.10

relative-Motion of two partiCles using translating axes

12–202. Determine the speed of the block at B.

6 m/s

99

12–205. If the end A of the cable is moving at vA = 3 m>s, determine the speed of block B. 12 A

D

C

vA  3 m/s

A B

B

Prob. 12–202

Prob. 12–205

12–203. If block A is moving downward with a speed of 2 m>s while C is moving up at 1 m>s, determine the speed of block B.

12–206. The motor draws in the cable at C with a constant velocity of vC = 4 m>s. The motor draws in the cable at D with a constant acceleration of aD = 8 m>s2. If vD = 0 when t = 0, determine (a) the time needed for block A to rise 3 m, and (b) the relative velocity of block A with respect to block B when this occurs.

C

D B

B C A

A Prob. 12–203

*12–204. Determine the speed of block A if the end of the rope is pulled down with a speed of 4 m>s.

Prob. 12–206 12–207. Determine the time needed for the load at B to attain a speed of 10 m>s, starting from rest, if the cable is drawn into the motor with an acceleration of 3 m>s2. *12–208. The cable at A is being drawn toward the motor at vA = 8 m>s. Determine the velocity of the block. A vA C

4 m/s B

B

A

Prob. 12–204

Probs. 12–207/208

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12–209. If the hydraulic cylinder H draws in rod BC at 12 1 m > s, determine the speed of slider A. A

*12–212. The roller at A is moving with a velocity of vA = 4 m>s and has an acceleration of aA = 2 m>s2 when x A = 3 m. Determine the velocity and acceleration of block B at this instant. vA

B

4 m/s

C

xA

H A

Prob. 12–209

4m

12–210. If the truck travels at a constant speed of vT = 1.8 m>s, determine the speed of the crate for any angle u of the rope. The rope has a length of 30 m and passes over a pulley of negligible size at A. Hint: Relate the coordinates xT and xC to the length of the rope and take the time derivative. Then substitute the trigonometric relation between xC and u. xC

xT

T

A 6m

C

B

Prob. 12–212

vT

12–213. The man pulls the boy up to the tree limb C by walking backward at a constant speed of 1.5 m>s. Determine the speed at which the boy is being lifted at the instant xA = 4 m. Neglect the size of the limb. When xA = 0, yB = 8 m, so that A and B are coincident, i.e., the rope is 16 m long.

u

Prob. 12–210 12–211. The crate C is being lifted by moving the roller at A downward with a constant speed of vA = 2 m>s along the guide. Determine the velocity and acceleration of the crate at the instant s = 1 m. When the roller is at B, the crate rests on the ground. Neglect the size of the pulley in the calculation. Hint: Relate the coordinates xC and xA using the problem geometry, then take the first and second time derivatives.

12–214. The man pulls the boy up to the tree limb C by walking backward. If he starts from rest when xA = 0 and moves backward with a constant acceleration aA = 0.2 m/s2, determine the speed of the boy at the instant yB = 4 m. Neglect the size of the limb. When xA = 0, yB = 8 m, so that A and B are coincident, i.e., the rope is 16 m long. C

yB

4m B

B xA

xC 4m

A

A C s

Prob. 12–211

xA

Probs. 12–213/214

8m

12.10

relative-Motion of two partiCles using translating axes

12–215. The motor draws in the cord at B with an acceleration of aB = 2 m>s2. When sA = 1.5 m, vB = 6 m>s. Determine the velocity and acceleration of the collar at this instant.

101

12–218. The boat can travel with a speed of 16 km>h in still water. The point of destination is located along the dashed 12 line. If the water is moving at 4 km>h, determine the bearing angle u at which the boat must travel to stay on course.

B vW  4 km/h

u 70

2m

A

sA

Prob. 12–218

Prob. 12–215 *12–216. If block B is moving down with a velocity vB and has an acceleration aB , determine the velocity and acceleration of block A in terms of the parameters shown.

12–219. The car is traveling at a constant speed of 100 km>h. If the rain is falling at 6 m>s in the direction shown, determine the velocity of the rain as seen by the driver.

sA A

h

30

vC

vr

100 km/h

vB, aB

Prob. 12–219

B

Prob. 12–216 12–217. At the instant shown, the car at A is traveling at 10 m>s around the curve while increasing its speed at 5 m>s2. The car at B is traveling at 18.5 m>s along the straightaway and increasing its speed at 2 m>s2. Determine the relative velocity and relative acceleration of A with respect to B at this instant.

*12–220. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 500 km>h and vB = 700 km>h such that the angle between their straightline courses is u = 60, determine the velocity of plane B with respect to plane A.

yB  18.5 m/s A

B

A 100 m

100 m 45

yA  10 m/s

vA  500 km/h

vB  700 km/h 60 B

Prob. 12–217

Prob. 12–220

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12–221. A car is traveling north along a straight road at 12 50 km>h. An instrument in the car indicates that the wind is coming from the east. If the car’s speed is 80 km>h, the instrument indicates that the wind is coming from the northeast. Determine the speed and direction of the wind. 12–222. Two boats leave the shore at the same time and travel in the directions shown. If vA = 10 m>s and vB = 15 m>s, determine the velocity of boat A with respect to boat B. How long after leaving the shore will the boats be 600 m apart?

*12–224. At the instant shown car A is traveling with a velocity of 30 m>s and has an acceleration of 2 m>s2 along the highway. At the same instant B is traveling on the trumpet interchange curve with a speed of 15 m>s, which is decreasing at 0.8 m>s2. Determine the relative velocity and relative acceleration of B with respect to A at this instant.

vA  10 m/s

A

A

vB  15 m/s

B

B

30

r

250 m

60 45

O

Prob. 12–224

Prob. 12–222

12–223. A man can row a boat at 5 m>s in still water. He wishes to cross a 50-m-wide river to point B, 50 m downstream. If the river flows with a velocity of 2 m>s, determine the speed of the boat and the time needed to make the crossing.

12–225. At the instant shown, car A has a speed of 20 km>h, which is being increased at the rate of 300 km>h2 as the car enters the expressway. At the same instant, car B is decelerating at 250 km>h2 while traveling forward at 100  km>h. Determine the velocity and acceleration of A with respect to B.

50 m 2 m/s A A 100 m

u

50 m

B

Prob. 12–223

B

Prob. 12–225

12.10

relative-Motion of two partiCles using translating axes

12–226. The man can row the boat in still water with a speed of 5 m>s. If the river is flowing at 2 m>s, determine the speed of the boat and the angle u he must direct the boat so that it travels from A to B.

103

*12–228. At the instant shown, cars A and B are traveling at velocities of 40 m>s and 30 m>s, respectively. If B is 12 increasing its velocity by 2 m>s2, while A maintains a constant velocity, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is rB = 200 m. 12–229. At the instant shown, cars A and B are traveling at velocities of 40 m>s and 30 m>s, respectively. If A is increasing its speed at 4 m>s2, whereas the speed of B is decreasing at 3 m>s2, determine the velocity and acceleration of B with respect to A. The radius of curvature at B is rB = 200 m.

B vw

5 m/s u

2 m/s

B

A

50 m

vA  40 m/s

vB  30 m/s 30

A 25 m

Prob. 12–226 Probs. 12–228/229

12–227. A passenger in an automobile observes that raindrops make an angle of 30° with the horizontal as the auto travels forward with a speed of 60 km>h. Compute the terminal (constant) velocity vr of the rain if it is assumed to fall vertically.

12–230. A man walks at 5 km>h in the direction of a 20 km>h wind. If raindrops fall vertically at 7 km>h in still air, determine direction in which the drops appear to fall with respect to the man.

vw  20 km/h

vr va  60 km/h

Prob. 12–227

vm  5 km/h

Prob. 12–230

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12–231. At the instant shown, car A travels along the 12 straight portion of the road with a speed of 25 m>s. At this same instant car B travels along the circular portion of the road with a speed of 15 m>s. Determine the velocity of car B relative to car A.

12–233. Car A travels along a straight road at a speed of 25 m>s while accelerating at 1.5 m>s2. At this same instant car C is traveling along the straight road with a speed of 30 m>s while decelerating at 3 m>s2. Determine the velocity and acceleration of car A relative to car C. 12–234. Car B is traveling along the curved road with a speed of 15 m>s while decreasing its speed at 2 m>s2. At this same instant car C is traveling along the straight road with a speed of 30 m>s while decelerating at 3 m>s2. Determine the velocity and acceleration of car B relative to car C.

15

15

45

1.5 m/s2 r

A

A

25 m/s

r  200 m

100 m 2 m/s2

30 B

3 m/s2

30

C

B

C

30 m/s

15 m/s

Prob. 12–231 Probs. 12–233/234

*12–232. At a given instant the football player at A throws a football C with a velocity of 20 m>s in the direction shown. Determine the constant speed at which the player at B must run so that he can catch the football at the same elevation at which it was thrown. Also calculate the relative velocity and relative acceleration of the football with respect to B at the instant the catch is made. Player B is 15 m away from A when A starts to throw the football.

12–235. The ship travels at a constant speed of vs = 20 m>s and the wind is blowing at a speed of vw = 10 m>s, as shown. Determine the magnitude and direction of the horizontal component of velocity of the smoke coming from the smoke stack as it appears to a passenger on the ship.

vs  20 m/s 30 vw  10 m/s

45 y

C

20 m/s A

x

60

B

15 m

Prob. 12–232

Prob. 12–235

12.10

relative-Motion of two partiCles using translating axes

105

ConCePtuAl ProblemS C12–1. If you measured the time it takes for the construction elevator to go from A to B, then B to C, and then C to D, and you also know the distance between each of the points, how could you determine the average velocity and average acceleration of the elevator as it ascends from A to D? Use numerical values to explain how this can be done.

12 C12–3. The basketball was thrown at an angle measured from the horizontal to the man’s outstretched arm. If the  basket is 3 m from the ground, make appropriate measurements in the photo and determine if the ball located as shown will pass through the basket.

D

C

B

Prob. C12–3 C12–4. The pilot tells you the wingspan of her plane and her constant airspeed. How would you determine the acceleration of the plane at the moment shown? Use numerical values and take any necessary measurements from the photo.

A

Prob. C12–1 C12–2. If the sprinkler at A is 1 m from the ground, then scale the necessary measurements from the photo to determine the approximate velocity of the water jet as it flows from the nozzle of the sprinkler.

A

Prob. C12–2

Prob. C12–4

106

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ChAPter reView Rectilinear Kinematics Rectilinear kinematics refers to motion along a straight line. A position coordinate s specifies the location of the particle on the line, and the displacement s is the change in this position.

s

O s

The average velocity is a vector quantity, defined as the displacement divided by the time interval.

vavg =

s

s t s s

O sT

The average speed is a scalar, and is the total distance traveled divided by the time of travel.

The time, position, velocity, and acceleration are related by three differential equations. If the acceleration is known to be constant, then the differential equations relating time, position, velocity, and acceleration can be integrated.

(vsp)avg =

a =

dv , dt

v =

ds , dt

sT t

a ds = v dv

v = v0 + a c t s = s0 + v0t + 12 act2 v2 = v20 + 2ac(s - s0)

Graphical Solutions If the motion is erratic, then it can be described by a graph. If one of these graphs is given, then the others can be established using the differential relations between a, v, s, and t.

dv , dt ds v = , dt a ds = v dv a =

107

Chapter review

Curvilinear Motion, x, y, z Curvilinear motion along the path can be resolved into rectilinear motion along the x, y, z axes. The equation of the path is used to relate the motion along each axis.

12 # vx = x

# a x = vx

# vy = y

# a y = vy

# vz = z

# a z = vz

z

s k i

v

a

z

r  xi  yj  zk

y

j

x

y

x

Projectile Motion Free-flight motion of a projectile follows a parabolic path. It has a constant velocity in the horizontal direction, and a constant downward acceleration of g = 9.81 m>s2 in the vertical direction. Any two of the three equations for constant acceleration apply in the vertical direction, and in the horizontal direction only one equation applies.

(+ c)

vy = (v0)y + act

(+ c)

y = y0 + (v0)yt + 12 act2

(+ c)

v2y = (v0)2y + 2ac(y - y0)

+ ) (S

x = x0 + (v0)xt

y

ag

vx v0

(v0)y

vy

(v0)x

v

r

y y0 x x0 x

108

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Curvilinear Motion n, t If normal and tangential axes are used for the analysis, then v is always in the positive t direction. The acceleration has two components. The tangential component, at, accounts for the change in the magnitude of the velocity; a slowing down is in the negative t direction, and a speeding up is in the positive t direction. The normal component an accounts for the change in the direction of the velocity. This component is always in the positive n direction.

Curvilinear Motion r, U If the path of motion is expressed in polar coordinates, then the velocity and acceleration components can be related to the time derivatives of r and u. To apply the time-derivative equations, # $ # $ it is necessary to determine r, r, r, u, u at the instant considered. If the path r = f(u) is given, then the chain rule of calculus must be used to obtain time derivatives. (See Appendix C.)

O¿ n

O

# at = v

or

an =

an

s

at ds = v dv v2 r

a at

v

t

# vr = r

v

# vu = ru # $ ar = r - ru2 $ # # au = ru + 2ru

vu vr P

r u O Velocity

Once the data are substituted into the equations, then the algebraic sign of the results will indicate the direction of the components of v or a along each axis.

a au ar r u O Acceleration

109

Chapter review

12

Absolute Dependent Motion of Two Particles The dependent motion of blocks that are suspended from pulleys and cables can be related by the geometry of the system. This is done by first establishing position coordinates, measured from a fixed origin to each block. Each coordinate must be directed along the line of motion of a block. Using geometry and/or trigonometry, the coordinates are then related to the cable length in order to formulate a position coordinate equation. The first time derivative of this equation gives a relationship between the velocities of the blocks, and a second time derivative gives the relation between their accelerations.

Datum sB

B

h

2sB + h + sA = l A

2vB = -vA 2aB = -aA

sA

Datum

Relative-Motion Analysis Using Translating Axes If two particles A and B undergo independent motions, then these motions can be related to their relative motion using a translating set of axes attached to one of the particles (A). For planar motion, each vector equation produces two scalar equations, one in the x, and the other in the y direction. For solution, the vectors can be expressed in Cartesian form, or the x and y scalar components can be written directly.

z¿

rB = rA + rB>A

a

z

vB = vA + vB>A aB = aA + aB>A

A a rA

Fixed observer

x¿

y¿

b B

y

O

x

Translating observer rB/A rB

b

110

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reView ProblemS R12–1. The position of a particle along a straight line is given by s = (t 3 - 9t 2 + 15t) m, where t is in seconds. Determine its maximum acceleration and maximum velocity during the time interval 0 … t … 10 s.

R12–2. If a particle has an initial velocity v0 = 12 m>s to the right, and a constant acceleration of 2 m>s2 to the left, determine the particle’s displacement in 10 s. Originally s0 = 0.

R12–5. A car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points A, B, and C. If it takes 3 s to go from A to B, and then 5 s to go from B to C, determine the average acceleration between points A and B and between points A and C.

y vC  40 m/s x

R12–3. A projectile, initially at the origin, moves along a straight-line path through a fluid medium such that its velocity is v = 1800(1 - e-0.3t) mm>s where t is in seconds. Determine the displacement of the projectile during the first 3 s.

vB  30 m/s B

C

45

vA  20 m/s A

Prob. R12–5

R12–4. The v–t graph of a car while traveling along a road is shown. Determine the acceleration when t = 2.5 s, 10 s, and 25 s. Also if s = 0 when t = 0, find the position when t = 5 s, 20 s, and 30 s.

R12–6. From a videotape, it was observed that a player kicked a football 40 m during a measured time of 3.6 seconds. Determine the initial speed of the ball and the angle u at which it was kicked.

v (m/s)

20 v0 u 5

20

Prob. R12–4

30

t (s)

A

40 m

Prob. R12–6

111

review probleMs R12–7. The truck travels in a circular path having a radius of 50 m at a speed of v = 4 m>s. For a short distance from # s = 0, its speed is increased by v = (0.05s) m>s2, where s is in meters. Determine its speed and the magnitude of its acceleration when it has moved s = 10 m.

R12–9. A particle is moving along a circular path of 2-m radius such that its position as a function of time is given by 12 u = (5t2) rad, where t is in seconds. Determine the magnitude of the particle’s acceleration when u = 30°. The particle starts from rest when u = 0°. R12–10. Determine the time needed for the load at B to attain a speed of 8 m>s, starting from rest, if the cable is drawn into the motor with an acceleration of 0.2 m>s2.

. v  (0.05s) m/s2 v  4 m/s A vA

50 m

Prob. R12–7 B

vB

Prob. R12–10 R12–8. Car B turns such that its speed is increased by (at)B = (0.5et) m>s2, where t is in seconds. If the car starts from rest when u = 0, determine the magnitudes of its velocity and acceleration when t = 2 s. Neglect the size of the car.

R12–11. Two planes, A and B, are flying at the same altitude. If their velocities are vA = 600 km>h and vB = 500 km>h such that the angle between their straightline courses is u = 75, determine the velocity of plane B with respect to plane A.

v B A vA

B

u 5m A

u

Prob. R12–8

vB

Prob. R12–11

Chapter 13

(© Migel/Shutterstock) A car driving along this road will be subjected to forces that create both normal and tangential accelerations. In this chapter we will study how these forces are related to the accelerations they create.

Kinetics of a Particle: Force and Acceleration Chapter ObjeCtives n

To state Newton’s Second Law of Motion and to define mass and weight.

n

To analyze the accelerated motion of a particle using the equation of motion with different coordinate systems.

n

To investigate central-force motion and apply it to problems in space mechanics.

13.1

Video Solutions are available for selected questions in this chapter.

Newton’s Second Law of Motion

Kinetics is a branch of dynamics that deals with the relationship between the change in motion of a body and the forces that cause this change. The basis for kinetics is Newton’s second law, which states that when an unbalanced force acts on a particle, the particle will accelerate in the direction of the force with a magnitude that is proportional to the force. This law can be verified experimentally by applying a known unbalanced force F to a particle, and then measuring the acceleration a. Since the force and acceleration are directly proportional, the constant of proportionality, m, may be determined from the ratio m = F>a. This positive scalar m is called the mass of the particle. Being constant during any acceleration, m provides a quantitative measure of the resistance of the particle to a change in its velocity, that is its inertia.

The jeep leans backward due to its inertia, which resists its forward acceleration.

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If the mass of the particle is m, Newton’s second law of motion may be written in mathematical form as F = ma 13

The above equation, which is referred to as the equation of motion, is one of the most important formulations in mechanics.* As previously stated, its validity is based solely on experimental evidence. In 1905, however, Albert Einstein developed the theory of relativity and placed limitations on the use of Newton’s second law for describing general particle motion. Through experiments it was proven that time is not an absolute quantity as assumed by Newton; and as a result, the equation of motion fails to predict the exact behavior of a particle, especially when the particle’s speed approaches the speed of light (0.3 Gm>s). Developments of the theory of quantum mechanics by Erwin Schrödinger and others indicate further that conclusions drawn from using this equation are also invalid when particles are the size of an atom and move close to one another. For the most part, however, these requirements regarding particle speed and size are not encountered in engineering problems, so their effects will not be considered in this book.

Newton’s Law of Gravitational Attraction. Shortly after formulating his three laws of motion, Newton postulated a law governing the mutual attraction between any two particles. In mathematical form this law can be expressed as F = G

m1 m2 r2

(13–1)

where F = force of attraction between the two particles G = universal constant of gravitation; according to experimental evidence G = 66.73(10-12) m3 >(kg # s2) m1 , m2 = mass of each of the two particles r = distance between the centers of the two particles

*Since m is constant, we can also write F = d(mv)>dt, where mv is the particle’s linear momentum. Here the unbalanced force acting on the particle is proportional to the time rate of change of the particle’s linear momentum.

13.1

newton’s seCond law of Motion

In the case of a particle located at or near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the particle. This force is termed the “weight” and, for our purpose, it will be the only gravitational force considered. From Eq. 13–1, we can develop a general expression for finding the weight W of a particle having a mass m1 = m. Let m2 = Me be the mass of the earth and r the distance between the earth’s center and the particle. Then, if g = GMe >r2, we have W = mg By comparison with F = ma, we term g the acceleration due to gravity. For most engineering calculations g is measured at a point on the surface of the earth at sea level, and at a latitude of 45°, which is considered the “standard location.” Here the values g = 9.81 m>s2 will be used for calculations. In the SI system the mass of the body is specified in kilograms, and the weight must be calculated using the above equation, Fig. 13–1. Thus,

W = mg (N)

( g = 9.81 m>s2)

(13–2)

As a result, a body of mass 1 kg has a weight of 9.81 N; a 2-kg body weighs 19.62 N; and so on.

m (kg)

a  g (m/s2)

W  mg (N)

Fig. 13–1

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The Equation of Motion

When more than one force acts on a particle, the resultant force is determined by a vector summation of all the forces; i.e., FR = F. For this more general case, the equation of motion may be written as

F1 (a)

13

of a

F = ma

To illustrate application of this equation, consider the particle shown in Fig. 13–2a, which has a mass m and is subjected to the action of two forces, F1 and F2 . We can graphically account for the magnitude and direction of each force acting on the particle by drawing the particle’s free-body diagram, Fig. 13–2b. Since the resultant of these forces produces the vector ma, its magnitude and direction can be represented graphically on the kinetic diagram, shown in Fig. 13–2c.* The equal sign written between the diagrams symbolizes the graphical equivalency between the free-body diagram and the kinetic diagram; i.e., F = ma.† In particular, note that if FR = F = 0, then the acceleration is also zero, so that the particle will either remain at rest or move along a straight-line path with constant velocity. Such are the conditions of static equilibrium, Newton’s first law of motion.

F2 FR  F

ma

 F1

Kinetic diagram

Free-body diagram (b)

Fig. 13–2

(c)

y a Path of particle vO O

Inertial frame of reference

Fig. 13–3

(13–3)

x

Inertial Reference Frame. When applying the equation of motion, it is important that the acceleration of the particle be measured with respect to a reference frame that is either fixed or translates with a constant velocity. In this way, the observer will not accelerate and measurements of the particle’s acceleration will be the same from any reference of this type. Such a frame of reference is commonly known as a Newtonian or inertial reference frame, Fig. 13–3. When studying the motions of rockets and satellites, it is justifiable to consider the inertial reference frame as fixed to the stars, whereas dynamics problems concerned with motions on or near the surface of the earth may be solved by using an inertial frame which is assumed fixed to the earth. Even though the earth both rotates about its own axis and revolves about the sun, the accelerations created by these rotations are relatively small and so they can be neglected for most applications. *Recall the free-body diagram considers the particle to be free of its surrounding supports and shows all the forces acting on the particle. The kinetic diagram pertains to the particle’s motion as caused by the forces. †The equation of motion can also be rewritten in the form F - ma = 0. The vector - ma is referred to as the inertia force vector. If it is treated in the same way as a “force vector,” then the state of “equilibrium” created is referred to as dynamic equilibrium. This method of application, which will not be used in this text, is often referred to as the D’Alembert principle, named after the French mathematician Jean le Rond d’Alembert.

13.2

the equation of Motion

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We are all familiar with the sensation one feels when sitting in a car that is subjected to a forward acceleration. Often people think this is caused by a “force” which acts on them and tends to push them back in their seats; however, this is not the case. Instead, this sensation occurs due to their inertia or the resistance of their mass to a change in velocity. Consider the passenger who is strapped to the seat of a rocket sled. Provided the sled is at rest or is moving with constant velocity, then no force is exerted on his back as shown on his free-body diagram.

Keystone/Hulton Archive/ Getty Images

13

Keystone/Hulton Archive/ Getty Images

When the thrust of the rocket engine causes the sled to accelerate, then the seat upon which he is sitting exerts a force F on him which pushes him forward with the sled. In the photo, notice that the inertia of his head resists this change in motion (acceleration), and so his head moves back against the seat and his face, which is nonrigid, tends to distort backward.

Keystone/Hulton Archive/ Getty Images

Upon deceleration the force of the seatbelt F tends to pull his body to a stop, but his head leaves contact with the back of the seat and his face distorts forward, again due to his inertia or tendency to continue to move forward. No force is pulling him forward, although this is the sensation he receives.

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Equation of Motion for a System of Particles

The equation of motion will now be extended to include a system of particles isolated within an enclosed region in space, as shown in Fig. 13–4a. In particular, there is no restriction in the way the particles are connected, so the following analysis applies equally well to the motion of a solid, liquid, or gas system. At the instant considered, the arbitrary i-th particle, having a mass mi , is subjected to a system of internal forces and a resultant external force. The internal force, represented symbolically as fi , is the resultant of all the forces the other particles exert on the ith particle. The resultant external force Fi represents, for example, the effect of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the system. The free-body and kinetic diagrams for the ith particle are shown in Fig. 13–4b. Applying the equation of motion, Fi + fi = miai

F = ma;

When the equation of motion is applied to each of the other particles of the system, similar equations will result. And, if all these equations are added together vectorially, we obtain Fi + fi = miai

z Fi

Fi fi

fi

i

ri



Free-body diagram

G rG

Kinetic diagram

y

x Inertial coordinate system (a)

(b)

Fig. 13–4

mi ai

13.3

equation of Motion for a systeM of partiCles

119

The summation of the internal forces, if carried out, will equal zero, since internal forces between any two particles occur in equal but opposite collinear pairs. Consequently, only the sum of the external forces will remain, and therefore the equation of motion, written for the system of particles, becomes 13

Fi = miai

(13–4)

If rG is a position vector which locates the center of mass G of the particles, Fig. 13–4a, then by definition of the center of mass, mrG = miri , where m = mi is the total mass of all the particles. Differentiating this equation twice with respect to time, assuming that no mass is entering or leaving the system, yields maG = miai Substituting this result into Eq. 13–4, we obtain

F = maG

(13–5)

Hence, the sum of the external forces acting on the system of particles is equal to the total mass of the particles times the acceleration of its center of mass G. Since in reality all particles must have a finite size to possess mass, Eq. 13–5 justifies application of the equation of motion to a body that is represented as a single particle.

important points • The equation of motion is based on experimental evidence and is • • • •

valid only when applied within an inertial frame of reference. The equation of motion states that the unbalanced force on a particle causes it to accelerate. An inertial frame of reference does not rotate, rather its axes either translate with constant velocity or are at rest. Mass is a property of matter that provides a quantitative measure of its resistance to a change in velocity. It is an absolute quantity and so it does not change from one location to another. Weight is a force that is caused by the earth’s gravitation. It is not absolute; rather it depends on the altitude of the mass from the earth’s surface.

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Equations of Motion: Rectangular Coordinates

When a particle moves relative to an inertial x, y, z frame of reference, the forces acting on the particle, as well as its acceleration, can be expressed in terms of their i, j, k components, Fig. 13–5. Applying the equation of motion, we have

13 z

F = ma;

For this equation to be satisfied, the respective i, j, k components on the left side must equal the corresponding components on the right side. Consequently, we may write the following three scalar equations:

Fz Fy Fx

z

y x

y

Fxi + Fy j + Fzk = m(axi + ay j + azk)

Fx = max Fy = may

(13–6)

Fz = maz

x

Fig. 13–5

In particular, if the particle is constrained to move only in the x–y plane, then the first two of these equations are used to specify the motion.

procedure for analysis The equations of motion are used to solve problems which require a  relationship between the forces acting on a particle and the accelerated motion they cause. Free-Body Diagram. • Select the inertial coordinate system. Most often, rectangular or x, y, z coordinates are chosen to analyze problems for which the particle has rectilinear motion. • Once the coordinates are established, draw the particle’s freebody diagram. Drawing this diagram is very important since it provides a graphical representation that accounts for all the forces (F) which act on the particle, and thereby makes it possible to resolve these forces into their x, y, z components. • The direction and sense of the particle’s acceleration a should also be established. If the sense is unknown, for mathematical convenience assume that the sense of each acceleration component acts in the same direction as its positive inertial coordinate axis. • The acceleration may be represented as the ma vector on the kinetic diagram.* • Identify the unknowns in the problem. *It is a convention in this text always to use the kinetic diagram as a graphical aid when developing the proofs and theory. The particle’s acceleration or its components will be shown as blue colored vectors near the free-body diagram in the examples.

13.4

equations of Motion: reCtangular Coordinates

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Equations of Motion. • If the forces can be resolved directly from the free-body diagram, apply the equations of motion in their scalar component form.

• If the geometry of the problem appears complicated, which often occurs in three dimensions, Cartesian vector analysis can be used for the solution.

• Friction. If a moving particle contacts a rough surface, it may be

necessary to use the frictional equation, which relates the frictional and normal forces Ff and N acting at the surface of contact by using the coefficient of kinetic friction, i.e., Ff = mkN. Remember that Ff always acts on the free-body diagram such that it opposes the motion of the particle relative to the surface it contacts. If the particle is on the verge of relative motion, then the coefficient of static friction should be used.

• Spring. If the particle is connected to an elastic spring having

negligible mass, the spring force Fs can be related to the deformation of the spring by the equation Fs = ks. Here k is the spring’s stiffness measured as a force per unit length, and s is the stretch or compression defined as the difference between the deformed length l and the undeformed length l0 , i.e., s = l - l0 .

Kinematics.

• If the velocity or position of the particle is to be found, it will be

necessary to apply the necessary kinematic equations once the particle’s acceleration is determined from F = ma.

• If acceleration is a function of time, use a = dv>dt and v = ds>dt

which, when integrated, yield the particle’s velocity and position, respectively.

• If acceleration is a function of displacement, integrate a ds = v dv to obtain the velocity as a function of position.

• If acceleration is constant, use v = v0 + ac t, s = s0 + v0t + 12 ac t2,

v2 = v20 + 2ac(s - s0) to determine the velocity or position of the particle.

• If the problem involves the dependent motion of several particles,

use the method outlined in Sec. 12.9 to relate their accelerations. In all cases, make sure the positive inertial coordinate directions used for writing the kinematic equations are the same as those used for writing the equations of motion; otherwise, simultaneous solution of the equations will result in errors.

• If the solution for an unknown vector component yields a negative scalar, it indicates that the component acts in the direction opposite to that which was assumed.

13

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13.1 P  400 N 30

13

Solution Using the equations of motion, we can relate the crate’s acceleration to the force causing the motion. The crate’s velocity can then be determined using kinematics. The weight of the crate is W = mg = Free-body Diagram. 50 kg (9.81 m>s2) = 490.5 N. As shown in Fig. 13–6b, the frictional force has a magnitude F = mkNC and acts to the left, since it opposes the motion of the crate. The acceleration a is assumed to act horizontally, in the positive x direction. There are two unknowns, namely NC and a. equations of Motion. Using the data shown on the free-body diagram, we have

(a)

y a x 490.5 N 400 N 30

F  0.3 NC NC (b)

The 50-kg crate shown in Fig. 13–6a rests on a horizontal surface for which the coefficient of kinetic friction is mk = 0.3. If the crate is subjected to a 400-N towing force as shown, determine the velocity of the crate in 3 s starting from rest.

+ Fx = max; S + c Fy = may;

400 cos 30 - 0.3NC = 50a NC - 490.5 + 400 sin 30 = 0

(1) (2)

Solving Eq. 2 for NC , substituting the result into Eq. 1, and solving for a yields NC = 290.5 N a = 5.185 m>s2 Kinematics. Notice that the acceleration is constant, since the applied force P is constant. Since the initial velocity is zero, the velocity of the crate in 3 s is + ) (S v = v0 + act = 0 + 5.185(3) = 15.6 m>s S Ans.

Fig. 13–6 490.5 N 400 N 30



50a

F  0.3NC NC

(c)

NOTE: We can also use the alternative procedure of drawing the crate’s free-body and kinetic diagrams, Fig. 13–6c, prior to applying the equations of motion.

13.4

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equations of Motion: reCtangular Coordinates

13.2

A 10-kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m>s, Fig. 13–7a. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected; and (b) atmospheric resistance is measured as FD = (0.01v2) N, where v is the speed of the projectile at any instant, measured in m>s.

z

Solution In both cases the known force on the projectile can be related to its acceleration using the equation of motion. Kinematics can then be used to relate the projectile’s acceleration to its position. (a)

part (a) Free-body Diagram. As shown in Fig. 13–7b, the projectile’s weight is W = mg = 10(9.81) = 98.1 N. We will assume the unknown acceleration a acts upward in the positive z direction. equation of Motion. + c Fz = maz; -98.1 = 10 a, a = -9.81 m>s2 The result indicates that the projectile, like every object having freeflight motion near the earth’s surface, is subjected to a constant downward acceleration of 9.81 m>s2. Kinematics. Initially, z0 = 0 and v0 = 50 m>s, and at the maximum height z = h, v = 0. Since the acceleration is constant, then (+ c ) v2 = v20 + 2 ac(z - z 0) 0 = (50)2 + 2(-9.81)(h - 0) Ans. h = 127 m part (b) Free-body Diagram. Since the force FD = (0.01v2) N tends to retard the upward motion of the projectile, it acts downward as shown on the free-body diagram, Fig. 13–7c. equation of Motion. + c Fz = maz; -0.01v2 - 98.1 = 10 a, a = -(0.001v2 + 9.81) Kinematics. Here the acceleration is not constant since FD depends on the velocity. Since a = f(v), we can relate a to position using (+ c ) a dz = v dv; -(0.001v2 + 9.81) dz = v dv Separating the variables and integrating, realizing that initially z0 = 0, v0 = 50 m>s (positive upward), and at z = h, v = 0, we have 0 0 v dv 2 2 = -500 ln(v dz = + 9810) 2 50 m>s L0 L50 m>s 0.001v + 9.81 h

h = 114 m

Ans.

NOTE: The answer indicates a lower elevation than that obtained in

part (a) due to atmospheric resistance or drag.

z

a 98.1 N (b)

z FD a 98.1 N (c)

Fig. 13–7

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A

B

C

The baggage truck A shown in the photo has a mass of 450 kg and tows a 275-kg cart B and a 160-kg cart C. For a short time the driving frictional force developed at the wheels of the truck is FA = (200t) N, where t is in seconds. If the truck starts from rest, determine its speed in 2 seconds. Also, what is the horizontal force acting on the coupling between the truck and cart B at this instant? Neglect the size of the truck and carts. 450 (9.81) N

275 (9.81) N

160 (9.81) N

FA

NC

NB

NA

(a)

Solution Free-body Diagram. As shown in Fig. 13–8a, it is the frictional driving force that gives both the truck and carts an acceleration. Here we have considered all three vehicles as a single system. equation of Motion. be considered. + Fx = max; d

Only motion in the horizontal direction has to 200t = (450 + 275 + 160)a a = 0.2260t

Kinematics. Since the acceleration is a function of time, the velocity of the truck is obtained using a = dv>dt with the initial condition that v0 = 0 at t = 0. We have L0

T

NA (b)

Fig. 13–8

dv =

L0

2s

0.2260t dt;

v = 0.1130t2 2

2s 0

= 0.452 m>s

Ans.

Free-body Diagram. In order to determine the force between the truck and cart B, we will consider a free-body diagram of the truck so that we can “expose” the coupling force T as external to the free-body diagram, Fig. 13–8b.

450 (9.81) N

FA

v

equation of Motion. When t = 2 s, then + Fx = max: d

200(2) - T = (450)[0.2260(2)] T = 197 N

Ans.

NOTE: Try and obtain this same result by considering a free-body diagram of carts B and C as a single system.

13.4

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A smooth 2-kg collar, shown in Fig. 13–9a, is attached to a spring having a stiffness k = 3 N>m and an unstretched length of 0.75 m. If the collar is released from rest at A, determine its acceleration and the normal force of the rod on the collar at the instant y = 1 m. Solution Free-body Diagram. The free-body diagram of the collar when it is located at the arbitrary position y is shown in Fig. 13–9b. Furthermore, the collar is assumed to be accelerating so that “a” acts downward in the positive y direction. There are four unknowns, namely, NC , Fs , a, and u. equations of Motion. + Fx = max; S + T Fy = may;

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equations of Motion: reCtangular Coordinates

-NC + Fs cos u = 0

(1)

19.62 - Fs sin u = 2a

(2)

From Eq. 2 it is seen that the acceleration depends on the magnitude and direction of the spring force. Solution for NC and a is possible once Fs and u are known. The magnitude of the spring force is a function of the stretch s of the spring; i.e., Fs = ks. Here the unstretched length is AB = 0.75 m, Fig. 13–9a; therefore, s = CB - AB = 2y2 + (0.75)2 - 0.75. Since k = 3 N>m, then Fs = ks = 3a 2y2 + (0.75)2 - 0.75b

(3)

From Fig. 13–9a, the angle u is related to y by trigonometry. tan u =

y 0.75

Substituting y = 1 m into Eqs. 3 and 4 yields Fs = 1.50 N and u = 53.1. Substituting these results into Eqs. 1 and 2, we obtain NC = 0.900 N a = 9.21 m>s2 T

Ans. Ans.

NOTE: This is not a case of constant acceleration, since the spring

force changes both its magnitude and direction as the collar moves downward.

0.75 m

B

A

13

u

y

k  3 N/m

C

(a)

x a

19.62 N

y (b)

Fig. 13–9

u NC

Fs

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Datum

13 sA C

sB

A B (a) T T

The 100-kg block A shown in Fig. 13–10a is released from rest. If the masses of the pulleys and the cord are neglected, determine the velocity of the 20-kg block B in 2 s. Solution Free-body Diagrams. Since the mass of the pulleys is neglected, then for pulley C, ma = 0 and we can apply Fy = 0, as shown in Fig. 13–10b. The free-body diagrams for blocks A and B are shown in Fig. 13–10c and d, respectively. Notice that for A to remain stationary T = 490.5 N, whereas for B to remain static T = 196.2 N. Hence A will move down while B moves up. Although this is the case, we will assume both blocks accelerate downward, in the direction of +sA and +sB . The three unknowns are T, aA , and aB . equations of Motion. Block A, + T Fy = may;

981 - 2T = 100aA

(1)

196.2 - T = 20aB

(2)

Block B, + T Fy = may; 2T (b)

2T

Kinematics. The necessary third equation is obtained by relating aA to aB using a dependent motion analysis, discussed in Sec. 12.9. The coordinates sA and sB in Fig. 13–10a measure the positions of A and B from the fixed datum. It is seen that 2sA + sB = l where l is constant and represents the total vertical length of cord. Differentiating this expression twice with respect to time yields 2aA = -aB

aA

sA 981 N (c)

T

aB

sB

196.2 N (d)

Fig. 13–10

(3)

Notice that when writing Eqs. 1 to 3, the positive direction was always assumed downward. It is very important to be consistent in this assumption since we are seeking a simultaneous solution of equations. The results are T = 327.0 N aA = 3.27 m>s2 aB = -6.54 m>s2 Hence when block A accelerates downward, block B accelerates upward as expected. Since aB is constant, the velocity of block B in 2 s is thus (+ T )

v = v0 + aBt = 0 + ( -6.54)(2) = -13.1 m>s The negative sign indicates that block B is moving upward.

Ans.

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127

preliMinary prObleMs P13–1. The 10-kg block is subjected to the forces shown. In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

P13–3. Determine the initial acceleration of the 10-kg smooth collar. The spring has an unstretched length of 1 m. 13

500 N 5

3 4

4m

300 N

3m

(a)

k  10 N/m

F  (20t) N

Prob. P13–3

(b)

Prob. P13–1

P13–2. The 10-kg block is subjected to the forces shown. In each case, determine its velocity at s = 8 m if v = 3 m>s at s = 0. Motion occurs to the right.

P13–4. Write the equations of motion in the x and y directions for the 10-kg block.

200 N

40 N

30 N y (a)

F  (2.5s) N

mk  0.2

30

(b)

Prob. P13–2

x

Prob. P13–4

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FunDaMental prObleMs F13–1. The motor winds in the cable with a constant acceleration, such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. Determine the tension 13 developed in the cable. The coefficient of kinetic friction between the crate and the plane is mk = 0.3.

F13–4. The 2-Mg car is being towed by a winch. If the winch exerts a force of T = 100(s + 1) N on the cable, where s is the displacement of the car in meters, determine the speed of the car when s = 10 m, starting from rest. Neglect rolling resistance of the car.

M s

Prob. F13–4

A

F13–5. The spring has a stiffness k = 200 N>m and is unstretched when the 25-kg block is at A. Determine the acceleration of the block when s = 0.4 m. The contact surface between the block and the plane is smooth.

30

Prob. F13–1

s

F13–2. If motor M exerts a force of F = (10t2 + 100) N on the cable, where t is in seconds, determine the velocity of the 25-kg crate when t = 4 s. The coefficients of static and kinetic friction between the crate and the plane are ms = 0.3 and mk = 0.25, respectively. The crate is initially at rest.

A

F  100 N

F  100 N

k  200 N/m

0.3 m M

Prob. F13–2 F13–3. A spring of stiffness k = 500 N>m is mounted against the 10-kg block. If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth.

F  500 N 5

3 4

Prob. F13–5 F13–6. Block B rests upon a smooth surface. If the coefficients of static and kinetic friction between A and B are ms = 0.4 and mk = 0.3, respectively, determine the acceleration of each block if P = 30 N.

A

P

s

10 kg

k  500 N/m 25 kg

B

Prob. F13–3

Prob. F13–6

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equations of Motion: reCtangular Coordinates

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prObleMs 13–1. The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. If the magnitude of P is increased until the crate begins to slide, determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction is mk = 0.3. 13–2. The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. Determine the crate’s acceleration in t = 2 s if the coefficient of static friction is ms = 0.4, the coefficient of kinetic friction is mk = 0.3, and the towing force is P = (90t2) N, where t is in seconds.

*13–4. If P = 400 N and the coefficient of kinetic friction between the 50-kg crate and the inclined plane is mk = 0.25, determine the velocity of the crate after it travels 6 m up the 13 plane.The crate starts from rest. 13–5. If the 50-kg crate starts from rest and travels a distance of 6 m up the plane in 4 s, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is mk = 0.25.

P 30°

p 20

30°

Probs. 13–1/2

Probs. 13–4/5

13–3. If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.

13–6. If the coefficient of kinetic friction between the 50-kg crate and the ground is mk = 0.3, determine the distance the crate travels and its velocity when t = 3 s. The crate starts from rest, and P = 200 N.

13–7. If the 50-kg crate starts from rest and achieves a velocity of v = 4 m>s when it travels a distance of 5 m to the right, determine the magnitude of force P acting on the crate. The coefficient of kinetic friction between the crate and the ground is mk = 0.3.

A

P

B 30 30

Prob. 13–3

Probs. 13–6/7

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*13–8. The conveyor belt is moving at 4 m>s. If the coefficient of static friction between the conveyor and the  10-kg package B is ms = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt.

and

a C C e l e r at i o n

13–11. Cylinder B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the block’s vertical position y so that when F is applied, the block rises with a constant acceleration aB. Neglect the mass of the cord and pulleys.

13 13–9. The conveyor belt is designed to transport packages of various weights. Each 10-kg package has a coefficient of kinetic friction mk = 0.15. If the speed of the conveyor is 5 m>s, and then it suddenly stops, determine the distance the package will slide on the belt before coming to rest.

d F y

B aB

Prob. 13–11

Probs. 13–8/9

13–10. The winding drum D is drawing in the cable at an accelerated rate of 5 m>s2. Determine the cable tension if the suspended crate has a mass of 800 kg.

B

*13–12. The elevator E has a mass of 500 kg and the counterweight at A has a mass of 150 kg. If the elevator attains a speed of 10 m>s after it rises 40 m, determine the constant force developed in the cable at B. Neglect the mass of the pulleys and cable.

D E B A

Prob. 13–10

Prob. 13–12

13.4 13–13. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = (3200t2) N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s when t = 0, determine its velocity when t = 2 s.

13–17. Determine the acceleration of the blocks when the system is released. The coefficient of kinetic friction is mk , and the mass of each block is m. Neglect the mass of the pulleys and cord.

13–14. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = (3200t2) N where t is in seconds. If the car has an initial velocity v1 = 2 m>s at s = 0 and t = 0, determine the distance it moves up the plane when t = 2 s.

13

B

M v1

131

equations of Motion: reCtangular Coordinates

A

2 m/s

Prob. 13–17

17

13–18. The motor lifts the 50-kg crate with an acceleration of 6 m>s2. Determine the components of force reaction and the couple moment at the fixed support A. 8

15

y

Probs. 13–13/14 13–15. The 75-kg man pushes on the 150-kg crate with a horizontal force F. If the coefficients of static and kinetic friction between the crate and the surface are ms = 0.3 and mk = 0.2, and the coefficient of static friction between the man’s shoes and the surface is ms = 0.8, show that the man is able to move the crate. What is the greatest acceleration the man can give the crate?

4m A

B

x 30

6 m/s2 F

Prob. 13–18

Prob. 13–15 *13–16. The 2-Mg truck is traveling at 15 m>s when the brakes on all its wheels are applied, causing it to skid for a distance of 10 m before coming to rest. Determine the constant horizontal force developed in the coupling C, and the frictional force developed between the tires of the truck and the road during this time. The total mass of the boat and trailer is 1 Mg.

13–19. A crate having a mass of 60 kg falls horizontally off  the back of a truck which is traveling at 80 km>h. Determine the coefficient of kinetic friction between the road and the crate if the crate slides 45 m on the ground with no tumbling along the road before coming to rest. Assume that the initial speed of the crate along the road is 80 km>h. 80 km/h

C

Prob. 13–16

Prob. 13–19

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*13–20. Determine the required mass of block A so that when it is released from rest it moves the 5-kg block B 0.75 m up along the smooth inclined plane in t = 2 s. Neglect the mass of the pulleys and cords. 13

and

a C C e l e r at i o n

13–23. The 50-kg block A is released from rest. Determine the velocity of the 15-kg block B in 2 s.

E

E C

C D B

D

A 60

B

Prob. 13–20 13–21. The force of the motor M on the cable is shown in the graph. Determine the velocity of the 400-kg crate A when t = 2 s.

A

F (N) 2500 F  625t 2

M

2

Prob. 13–23 t (s)

*13–24. If the supplied force F = 150 N, determine the velocity of the 50-kg block A when it has risen 3 m, starting from rest.

A

Prob. 13–21 13–22. The bullet of mass m is given a velocity due to gas pressure caused by the burning of powder within the chamber of the gun. Assuming this pressure creates a force of F = F0 sin (pt>t0) on the bullet, determine the velocity of the bullet at any instant it is in the barrel. What is the bullet’s maximum velocity? Also, determine the position of the bullet in the barrel as a function of time.

C B

F

F

F0

A t0

Prob. 13–22

t

Prob. 13–24

13.4 13–25. A 60-kg suitcase slides from rest 5 m down the smooth ramp. Determine the distance R where it strikes the ground at B. How long does it take to go from A to B? 13–26. Solve Prob. 13–25 if the suitcase has an initial velocity down the ramp of vA = 2 m>s, and the coefficient of kinetic friction along AC is mk = 0.2.

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equations of Motion: reCtangular Coordinates

13–29. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt. 13

A 4 m/s B

5m

30

C

30

2.5 m B

Prob. 13–29 R

Probs. 13–25/26 13–27. The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Take u = 30.

13–30. The 1.5 Mg sports car has a tractive force of F = 4.5 kN. If it produces the velocity described by v-t graph shown, plot the air resistance R versus t for this time period.

*13–28. The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the smallest incline u of the ramp so that the crates will slide off and fall into the cart.

R

F v (m/s)

vA  2.5 m/s

v  (–0.05t 2 + 3t ) m/s 45

A 3m

u

B

30

Probs. 13–27/28

Prob. 13–30

t (s)

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13–31. Crate B has a mass m and is released from rest when it is on top of cart A, which has a mass 3m. Determine the tension in cord CD needed to hold the cart from moving while B is sliding down A. Neglect friction. 13

B

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a C C e l e r at i o n

13–34. The 300-kg bar B, originally at rest, is being towed over a series of small rollers. Determine the force in the cable when t = 5 s, if the motor M is drawing in the cable for a short time at a rate of v = (0.4t2) m>s, where t is in seconds (0 … t … 6 s). How far does the bar move in 5 s? Neglect the mass of the cable, pulley, and the rollers.

M D

A

C u

Prob. 13–31 *13–32. The 4-kg smooth cylinder is supported by the spring having a stiffness of kAB = 120 N>m. Determine the velocity of the cylinder when it moves downward s = 0.2 m from its equilibrium position, which is caused by the application of the force F = 60 N.

v B

F  60 N

Prob. 13–34 B s kAB  120 N/m

A

13–35. An electron of mass m is discharged with an initial horizontal velocity of v0. If it is subjected to two fields of force for which Fx = F0 and Fy = 0.3F0, where F0 is constant, determine the equation of the path, and the speed of the electron at any time t.

y

Prob. 13–32

Prob. 13–33

++++++++++++++ ++++++++++++++

13–33. The coefficient of static friction between the 200-kg crate and the flat bed of the truck is ms = 0.3. Determine the shortest time for the truck to reach a speed of 60 km>h, starting from rest with constant acceleration, so that the crate does not slip.

v0

Prob. 13–35

x

13.4 *13–36. A car of mass m is traveling at a slow velocity v0. If it is subjected to the drag resistance of the wind, which is proportional to its velocity, i.e., FD = kv, determine the distance and the time the car will travel before its velocity becomes 0.5v0. Assume no other frictional forces act on the car.

equations of Motion: reCtangular Coordinates

135

13–38. Blocks A and B each has a mass m. Determine the largest horizontal force P which can be applied to B so that it will not slide on A. Also, what is the corresponding acceleration? The coefficient of static friction between A and B is ms. Neglect any friction between A and the horizontal surface.

P

B

v

FD

A



Prob. 13–36 Prob. 13–38

13–37. The 10-kg block A rests on the 50-kg plate B in the position shown. Neglecting the mass of the rope and pulley, and using the coefficients of kinetic friction indicated, determine the time needed for block A to slide 0.5 m on the plate when the system is released from rest.

13–39. The tractor is used to lift the 150-kg load B with the 24-m-long rope, boom, and pulley system. If the tractor travels to the right at a constant speed of 4 m>s, determine the tension in the rope when sA = 5 m. When sA = 0, sB = 0. *13–40. The tractor is used to lift the 150-kg load B with the 24-m-long rope, boom, and pulley system. If the tractor travels to the right with an acceleration of 3 m>s2 and has a velocity of 4 m>s at the instant sA = 5 m, determine the tension in the rope at this instant. When sA = 0, sB = 0.

12 m 0.5 m

mAB  0.2

C

A B

sB

B

mBC  0.1

30

A sA

Prob. 13–37

Probs. 13–39/40

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13–41. A freight elevator, including its load, has a mass of 1  Mg. It is prevented from rotating due to the track and wheels mounted along its sides. If the motor M develops a constant tension T = 4 kN in its attached cable, determine the velocity of the elevator when it has moved upward 6 m starting from rest. Neglect the mass of the pulleys and cables.

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13–43. If the force exerted on cable AB by the motor is F = ( 100t 3>2 ) N, where t is in seconds, determine the 50-kg crate’s velocity when t = 5 s. The coefficients of static and kinetic friction between the crate and the ground are ms = 0.4 and mk = 0.3, respectively. Initially the crate is at rest.

A

B

M

Prob. 13–43

*13–44. A parachutist having a mass m opens his parachute from an at-rest position at a very high altitude. If the atmospheric drag resistance is FD = kv 2, where k is a constant, determine his velocity when he has fallen for a time t. What is his velocity when he lands on the ground? This velocity is referred to as the terminal velocity, which is found by letting the time of fall t S  .

FD

Prob. 13–41 13–42. If the motor draws in the cable with an acceleration of 3 m>s2, determine the reactions at the supports A and B. The beam has a uniform mass of 30 kg>m, and the crate has a mass of 200 kg. Neglect the mass of the motor and pulleys. 2.5 m

0.5 m

v

3m

A

B

Prob. 13–44 3 m/s2

13–45. Each of the three plates has a mass of 10 kg. If the coefficients of static and kinetic friction at each surface of contact are ms = 0.3 and mk = 0.2, respectively, determine the acceleration of each plate when the three horizontal forces are applied. 18 N

D C

C 15 N

B A

Prob. 13–42

Prob. 13–45

100 N

13.4 13–46. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

13–47. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip on B. The coefficient of static friction between A and B is ms. Neglect any friction between B and C.

A

equations of Motion: reCtangular Coordinates

137

13–49. Block A has a mass m A and is attached to a spring having a stiffness k and unstretched length l0. If another block B, having a mass m B , is pressed against A so that the spring deforms a distance d, determine the distance both blocks slide on the smooth surface before they begin to separate. What is their velocity at this instant?

13–50. Block A has a mass m A and is attached to a spring having a stiffness k and unstretched length l0. If another block B, having a mass m B , is pressed against A so that the spring deforms a distance d, show that for separation to occur it is necessary that d 7 2mk g ( m A + m B ) >k, where mk is the coefficient of kinetic friction between the blocks and the ground. Also, what is the distance the blocks slide on the surface before they separate?

P

u B C

k

Probs. 13–46/47

A

B

Probs. 13–49/50

*13–48. The smooth block B of negligible size has a mass m and rests on the horizontal plane. If the board AC pushes on the block at an angle u with a constant acceleration a0, determine the velocity of the block along the board and the distance s the block moves along the board as a function of time t. The block starts from rest when s = 0, t = 0.

13–51. The block A has a mass m A and rests on the pan B, which has a mass m B . Both are supported by a spring having a stiffness k that is attached to the bottom of the pan and to the ground. Determine the distance d the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched.

A

a0 C

u A

B

B y d

k

s

Prob. 13–48

Prob. 13–51

13

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Equations of Motion: Normal and Tangential Coordinates

When a particle moves along a curved path which is known, the equation of motion for the particle may be written in the tangential, normal, and binormal directions, Fig. 13–11. Note that there is no motion of the particle in the binormal direction, since the particle is constrained to move along the path. We have

13

F = ma

b

Ftut + Fnun + Fbub = mat + man

Fbub

O

t n

Fnun

Ftut P

Inertial coordinate system

Fig. 13–11

This equation is satisfied provided Ft = mat (13–7)

Fn = man Fb = 0

Recall that at (= dv>dt) represents the time rate of change in the magnitude of velocity. So if Ft acts in the direction of motion, the particle’s speed will increase, whereas if it acts in the opposite direction, the particle will slow down. Likewise, an (= v2 >r) represents the time rate of change in the velocity’s direction. It is caused by Fn , which always acts in the positive n direction, i.e., toward the path’s center of curvature. For this reason it is often referred to as the centripetal force.

n t

As a roller coaster falls downward along the track, the cars have both a normal and a tangential component of acceleration.

13.5

139

equations of Motion: norMal and tangential Coordinates

procedure for analysis When a problem involves the motion of a particle along a known curved path, normal and tangential coordinates should be considered for the analysis since the acceleration components can be readily formulated. The method for applying the equations of motion, which relate the forces to the acceleration, has been outlined in the procedure given in Sec. 13.4. Specifically, for t, n, b coordinates it may be stated as follows: Free-Body Diagram.

• Establish the inertial t, n, b coordinate system at the particle and draw the particle’s free-body diagram.

• The particle’s normal acceleration an always acts in the positive n direction.

• If the tangential acceleration at is unknown, assume it acts in the positive t direction.

• There is no acceleration in the b direction. • Identify the unknowns in the problem. Equations of Motion.

• Apply the equations of motion, Eq. 13–7. Kinematics.

• Formulate the tangential and normal components of acceleration; i.e., at = dv>dt or at = v dv>ds and an = v2 >r.

• If the path is defined as y = f(x), the radius of curvature at the

point where the particle is located can be obtained from r = [1 + (dy>dx)2]3>2 >  d2y>dx2  .

b

n

t

The unbalanced force of the rope on the skier gives him a normal component of acceleration.

13

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13.6

Determine the banking angle u for the race track so that the wheels of the racing cars shown in Fig. 13–12a will not have to depend upon friction to prevent any car from sliding up or down the track. Assume the cars have negligible size, a mass m, and travel around the curve of radius r with a constant speed v.

u

(a)

Solution Before looking at the following solution, give some thought as to why it should be solved using t, n, b coordinates. Free-body Diagram. As shown in Fig. 13–12b, and as stated in the problem, no frictional force acts on the car. Here NC represents the resultant of the ground on all four wheels. Since an can be calculated, the unknowns are NC and u. equations of Motion.

+ c Fb = 0;

NC sin u = m NC cos u - mg = 0

v2 r

v2 b gr

u W  mg (b)

(2)

Fig. 13–12

v2 gr

u = tan-1 a

n

(1)

Eliminating NC and m from these equations by dividing Eq. 1 by Eq. 2, we obtain tan u =

an

NC

Using the n, b axes shown,

+ F = ma ; S n n

b

Ans.

NOTE: The result is independent of the mass of the car. Also, a force

summation in the tangential direction is of no consequence to the solution. If it were considered, then at = dv>dt = 0, since the car moves with constant speed. A further analysis of this problem is discussed in Prob. 21–52.

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141

13.7 The 3-kg disk D is attached to the end of a cord as shown in Fig. 13–13a. The other end of the cord is attached to a ball-and-socket joint located at the center of a platform. If the platform rotates rapidly, and the disk is placed on it and released from rest as shown, determine the time it takes for the disk to reach a speed great enough to break the cord. The maximum tension the cord can sustain is 100 N, and the coefficient of kinetic friction between the disk and the platform is mk = 0.1. Motion of platform

D

1m

(a)

Solution Free-body Diagram. The frictional force has a magnitude F = mkND = 0.1ND and a sense of direction that opposes the relative motion of the disk with respect to the platform. It is this force that gives the disk a tangential component of acceleration causing v to increase, thereby causing T to increase until it reaches 100 N. The weight of the disk is W = 3(9.81) N = 29.43 N. Since an can be related to v, the unknowns are ND , at , and v.

b

29.43 N T

F  0.1ND t

at

ND (b)

Fig. 13–13

an

n

equations of Motion. Fn = man; Ft = mat; Fb = 0;

T = 3a 0.1ND = 3at

v2 b 1

ND - 29.43 = 0

(1) (2) (3)

Setting T = 100 N, Eq. 1 can be solved for the critical speed vcr of the disk needed to break the cord. Solving all the equations, we obtain ND = 29.43 N at = 0.981 m>s2 vcr = 5.77 m>s Kinematics. Since at is constant, the time needed to break the cord is vcr = v0 + att 5.77 = 0 + (0.981)t t = 5.89 s

Ans.

13

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Design of the ski jump shown in the photo requires knowing the type of forces that will be exerted on the skier and her approximate trajectory. If in this case the jump can be approximated by the parabola shown in Fig. 13–14a, determine the normal force on the 70-kg skier the instant she arrives at the end of the jump, point A, where her velocity is 20 m>s. Also, what is her acceleration at this point? Solution Why consider using n, t coordinates to solve this problem? Free-body Diagram. Since dy>dx = x>30 0 x = 0 = 0, the slope at A is horizontal. The free-body diagram of the skier when she is at A is shown in Fig. 13–14b. Since the path is curved, there are two components of acceleration, an and at . Since an can be calculated, the unknowns are at and NA .

NA - 70(9.81) = 70 c

+ Ft = mat; d

0 = 70at

(20)2 d r

[1 + (dy>dx)2]3>2 2

2

 d y>dx 

x 60 m

A

n

(2)

2 x=0

=

[1 + (0)2]3>2 1  30 

an 70(9.81) N

t

at

= 30 m (b)

NA

Substituting this into Eq. 1 and solving for NA , we obtain NA = 1620 N Kinematics.

1 x2  60 60

(1)

The radius of curvature r for the path must be determined at point 1 2 1 1 A(0, -60 m). Here y = 60 x - 60, dy>dx = 30 x, d2y>dx2 = 30 , so that at x = 0, r =

y

(a)

equations of Motion. + c Fn = man;

y

Ans.

From Eq. 2, at = 0

Thus, an =

(20)2 v2 = = 13.33 m>s2 r 30

aA = an = 13.3 m>s2 c

Ans.

NOTE: Apply the equation of motion in the y direction and show that when the skier is in midair, her downward acceleration is 9.81 m > s2.

Fig. 13–14

13.5

equations of Motion: norMal and tangential Coordinates

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13.9

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The 60-kg skateboarder in Fig. 13–15a coasts down the circular track. If he starts from rest when u = 0, determine the magnitude of the normal reaction the track exerts on him when u = 60. Neglect his size for the calculation.

O u 4m

Solution Free-body Diagram. The free-body diagram of the skateboarder when he is at an arbitrary position u is shown in Fig. 13–15b. At u = 60 there are three unknowns, Ns , at , and an (or v).

(a) 60 (9.81) N n

equations of Motion.

u

+ Q Fn = man; Ns - [60(9.81)N] sin u = (60 kg)a

an

+ R Ft = mat;

[60(9.81)N] cos u = (60 kg) at

at

v2 b 4m

(1)

at = 9.81 cos u

Ns

Kinematics. Since at is expressed in terms of u, the equation v dv = at ds must be used to determine the speed of the skateboarder when u = 60. Using the geometric relation s = ur, where ds = r du = (4 m) du, Fig. 13–15c, and the initial condition v = 0 at u = 0, we have,

t (b)

v dv = at ds O u 4m

du

ds  4du

L0

v

v dv = 2 v

L0

60

9.81 cos u(4 du)

60 v 2 = 39.24 sin u 2 2 0 0

v2 - 0 = 39.24(sin 60 - 0) 2 (c)

Fig. 13–15

v2 = 67.97 m2 >s2

Substituting this result and u = 60 into Eq. (1), yields Ns = 1529.23 N = 1.53 kN

Ans.

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preliMinary prObleMs

13

P13–5. Set up the n, t axes and write the equations of motion for the 10-kg block along each of these axes.

P13–6. Set up the n, b, t axes and write the equations of motion for the 10-kg block along each of these axes.

6 m/s

mk  0.3 10 m

4m mk  0.2 8 m/s

(a) (a)

30 5m 4 m/s v

2m

mk  0.2 (b)

ms  0.3

60 6m

8 m/s (c)

Prob. P13–5

Constant rotation Block has impending motion

(b)

Prob. P13–6

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FunDaMental prObleMs F13–7. The block rests at a distance of 2 m from the center of the platform. If the coefficient of static friction between the block and the platform is ms = 0.3, determine the maximum speed which the block can attain before it begins to slip. Assume the angular motion of the disk is slowly increasing. z

F13–10. The sports car is traveling along a 30 banked road having a radius of curvature of r = 150 m. If the coefficient of static friction between the tires and the road is ms = 0.2, determine the maximum safe speed so no slipping 13 occurs. Neglect the size of the car. r  150 m

u  30

2m

Prob. F13–10

Prob. F13–7 F13–8. Determine the maximum speed that the jeep can travel over the crest of the hill and not lose contact with the road.

F13–11. If the 10-kg ball has a velocity of 3 m>s when it is at the position A, along the vertical path, determine the tension in the cord and the increase in the speed of the ball at this position.

2m

r  75 m

O

u  45

A

Prob. F13–8 F13–9. A pilot weighs 70 kg and is traveling at a constant speed of 36 m>s. Determine the normal force he exerts on the seat of the plane when he is upside down at A. The loop has a radius of curvature of 120 m. A

Prob. F13–11 F13–12. The motorcycle has a mass of 0.5 Mg and a negligible size. It passes point A traveling with a speed of 15 m>s, which is increasing at a constant rate of 1.5 m>s2. Determine the resultant frictional force exerted by the road on the tires at this instant.

120 m

rA  200 m

A

Prob. F13–9

3 m/s

Prob. F13–12

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prObleMs *13–52. A girl having a mass of 25 kg sits at the edge of the merry-go-round so her center of mass G is at a distance of 13 1.5 m from the axis of rotation. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which she can have before she begins to slip off the merry-go-round. The coefficient of static friction between the girl and the merry-go-round is ms = 0.3.

13–54. The collar A, having a mass of 0.75 kg, is attached to a spring having a stiffness of k = 200 N>m. When rod BC rotates about the vertical axis, the collar slides outward along the smooth rod DE. If the spring is unstretched when s = 0, determine the constant speed of the collar in order that s = 100 mm. Also, what is the normal force of the rod on the collar? Neglect the size of the collar.

z

C

1.5 m

k  200 N/m D A

G

E s B

Prob. 13–52 13–53. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5 m from the platform’s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is m = 0.2. z

Prob. 13–54 13–55. A 5-Mg airplane is flying at a constant speed of 350  km>h along a horizontal circular path of radius r = 3000 m. Determine the uplift force L acting on the airplane and the banking angle u. Neglect the size of the airplane. *13–56. A 5-Mg airplane is flying at a constant speed of 350 km>h along a horizontal circular path. If the banking angle u = 15°, determine the uplift force L acting on the airplane and the radius r of the circular path. Neglect the size of the airplane.

5m L u

r

Prob. 13–53

Probs. 13–55/56

13.5

equations of Motion: norMal and tangential Coordinates

13–57. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block is given a speed of v = 10 m>s, determine the radius r of the circular path along which it travels. 13–58. The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through a hole in the center of the smooth table. If the block travels along a circular path of radius r = 1.5 m, determine the speed of the block.

147

*13–60. Determine the maximum constant speed at which the pilot can travel around the vertical curve having a radius of curvature r = 800 m, so that he experiences a maximum acceleration an = 8g = 78.5 m>s2. If he has a mass of 70 kg, determine the normal force he exerts on the seat of the airplane when the plane is traveling at this speed and is at 13 its lowest point.

r r  800 m B v

A

Prob. 13–60 Probs. 13–57/58

13–59. Cartons having a mass of 5 kg are required to move along the assembly line at a constant speed of 8 m>s. Determine the smallest radius of curvature, r, for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are ms = 0.7 and mk = 0.5, respectively.

13–61. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the minimum constant speed the spool can have so that it does not slip down the rod. 13–62. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the maximum constant speed the spool can have so that it does not slip up the rod. z 5 4

8 m/s

S 0.25 m

r A

Prob. 13–59

Probs. 13–61/62

3

148

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13–63. The 1.40-Mg helicopter is traveling at a constant speed of 40 m>s along the horizontal curved path while banking at u = 30°. Determine the force acting normal to the blade, i.e., in the y¿ direction, and the radius of curvature of the path. 13 *13–64. The 1.40-Mg helicopter is traveling at a constant speed of 33 m>s along the horizontal curved path having a radius of curvature of r = 300 m. Determine the force the blade exerts on the frame and the bank angle u.

and

a C C e l e r at i o n

13–66. Determine the constant speed of the passengers on the amusement-park ride if it is observed that the supporting cables are directed at u = 30 from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what are the components of force in the n, t, and b directions which the chair exerts on a 50-kg passenger during the motion?

4m y y¿

u

b u

6m t

n r

Probs. 13–63/64

13–65. The vehicle is designed to combine the feel of a motorcycle with the comfort and safety of an automobile. If the vehicle is traveling at a constant speed of 80 km>h along a circular curved road of radius 100 m, determine the tilt angle u of the vehicle so that only a normal force from the seat acts on the driver. Neglect the size of the driver.

Prob. 13–66

13–67. Bobs A and B of mass mA and mB (mA > mB) are connected to an inextensible light string of length l that passes through the smooth ring at C. If bob B moves as a conical pendulum such that A is suspended a distance of h from C, determine the angle u and the speed of bob B. Neglect the size of both bobs.

u C

h

u

A B

Prob. 13–65

Prob. 13–67

149

equations of Motion: norMal and tangential Coordinates

13.5

*13–68. Prove that if the block is released from rest at point B of a smooth path of arbitrary shape, the speed it attains when it reaches point A is equal to the speed it attains when it falls freely through a distance h; i.e., v = 22gh.

13–71. A ball having a mass 2 kg and negligible size moves within a smooth vertical circular slot. If it is released from rest when u = 10, determine the force of the slot on the ball when the ball arrives at points A and B. 13

B 10

h

u A

0.8 m

r

A

Prob. 13–68 13–69. The skier starts from rest at A(10 m, 0) and descends the smooth slope, which may be approximated by a parabola. If she has a mass of 52 kg, determine the normal force the ground exerts on the skier at the instant she arrives at point B. Neglect the size of the skier.

10 B

Prob. 13–71

y 10 m

y

1 2 –– x 20

*13–72. The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m>s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car.

5 A

x

5m B

Prob. 13–69 13–70. The 800-kg motorbike travels with a constant speed of 80 km>h up the hill. Determine the normal force the surface exerts on its wheels when it reaches point A. Neglect its size.

13–73. The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m>s and increasing its speed at 3 m>s2. Determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at this instant. Neglect the size of the car. y

y A

y  20 (1  y2

2x

A

x 100 m

Prob. 13–70

x2 ) 6400

80 m

Probs. 13–72/73

x

150

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partiCle: forCe

13–74. The block B, having a mass of 0.2 kg, is attached to the vertex A of the right circular cone using a light cord. The cone is rotating at a constant angular rate about the z axis such that the block attains a speed of 0.5 m>s. At this speed, determine the tension in the cord and the reaction which the cone exerts on the block. Neglect the size of the block and 13 the effect of friction.

and

a C C e l e r at i o n

*13–76. The 35-kg box has a speed 2 m>s when it is at A on the smooth ramp. If the surface is in the shape of a parabola, determine the normal force on the box at the instant x = 3 m. Also, what is the rate of increase in its speed at this instant? y

2 m/s

A

z

y

4

1 –– x2 9

A

200 mm

B

x

Prob. 13–76

400 mm

13–77. The box has a mass m and slides down the smooth chute having the shape of a parabola. If it has an initial velocity of v0 at the origin, determine its velocity as a function of x. Also, what is the normal force on the box, and the tangential acceleration as a function of x?

300 mm

y

Prob. 13–74 x x

13–75. Determine the maximum speed at which the car with mass m can pass over the top point A of the vertical curved road and still maintain contact with the road. If the car maintains this speed, what is the normal reaction the road exerts on the car when it passes the lowest point B on the road?

y  0.5x2

Prob. 13–77

r

A

r

13–78. Determine the maximum constant speed at which the 2-Mg car can travel over the crest of the hill at A without leaving the surface of the road. Neglect the size of the car in the calculation. y 2

B r

A

r

y  20 (1  x ) 10 000

x 100 m

Prob. 13–75

Prob. 13–78

13.5

equations of Motion: norMal and tangential Coordinates

13–79. A collar having a mass 0.75 kg and negligible size slides over the surface of a horizontal circular rod for which the coefficient of kinetic friction is mk = 0.3. If the collar is given a speed of 4 m>s and then released at u = 0, determine how far, s, it slides on the rod before coming to rest. z

100 mm

151

13–83. The airplane, traveling at a constant speed of 50 m>s, is executing a horizontal turn. If the plane is banked at u = 15, when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature r of the turn. Also, what is the normal force of the seat on the pilot if he has a mass of 70 kg.

u y u s

r v

x

Prob. 13–79 *13–80. The 8-kg sack slides down the smooth ramp. If it has a speed of 1.5 m>s when y = 0.2 m, determine the normal reaction the ramp exerts on the sack and the rate of increase in the speed of sack at this instant.

Prob. 13–83

y

y  0.2ex

x

*13–84. The ball has a mass m and is attached to the cord of length l. The cord is tied at the top to a swivel and the ball is given a velocity v0. Show that the angle u which the cord makes with the vertical as the ball travels around the circular path must satisfy the equation tan u sin u = v02 >gl. Neglect air resistance and the size of the ball.

Prob. 13–80 13–81. The 2-kg pendulum bob moves in the vertical plane with a velocity of 8 m>s when u = 0. Determine the initial tension in the cord and also at the instant the bob reaches u = 30. Neglect the size of the bob.

O

13–82. The 2-kg pendulum bob moves in the vertical plane with a velocity of 6 m>s when u = 0. Determine the angle u where the tension in the cord becomes zero.

u l



2m

Probs. 13–81/82

v0

Prob. 13–84

13

152

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Fzuz

partiCle: forCe

13.6

P

Frur

a C C e l e r at i o n

Equations of Motion: Cylindrical Coordinates

Fuuu

13

and

When all the forces acting on a particle are resolved into cylindrical components, i.e., along the unit-vector directions ur , uu , uz , Fig. 13–16, the equation of motion can be expressed as

z

O

F = ma

u

Frur + Fuuu + Fzuz = marur + mauuu + mazuz

r

Inertial coordinate system

To satisfy this equation, we require Fr = mar

Fig. 13–16

(13–8)

Fu = mau Fz = maz

If the particle is constrained to move only in the r9u plane, then only the first two of Eq. 13–8 are used to specify the motion.

Tangential and Normal Forces. The most straightforward type of problem involving cylindrical coordinates requires the determination of the resultant force components Fr , Fu , Fz which cause a particle to move with a known acceleration. If, however, the particle’s accelerated motion is not completely specified at the given instant, then some information regarding the directions or magnitudes of the forces acting on the particle must be known or calculated in order to solve Eq. 13–8. For example, the force P causes the particle in Fig. 13–17a to move along a path r = f(u). The normal force N which the path exerts on the particle is always perpendicular to the tangent of the path, whereas the frictional force F always acts along the tangent in the opposite direction of motion. The directions of N and F can be specified relative to the radial coordinate by using the angle c, Fig. 13–17b, which is defined between the extended radial line and the tangent to the curve. r  f (u)

r  f (u)

Tangent

Tangent

c N

r O

Motion of the roller coaster along this spiral can be studied using cylindrical coordinates.

r

F

u P

O (a)

u (b)

Fig. 13–17

13.6

This angle can be obtained by noting that when the particle is displaced a distance ds along the path, Fig. 13–17c, the component of displacement in the radial direction is dr and the component of displacement in the transverse direction is r du. Since these two components are mutually perpendicular, the angle c can be determined from tan c = r du>dr, or tan c =

r dr>du

153

equations of Motion: CylindriCal Coordinates r  f (u) Tangent dr r du du

(13–9)

If c is calculated as a positive quantity, it is measured from the extended radial line to the tangent in a counterclockwise sense or in the positive direction of u. If it is negative, it is measured in the opposite direction to positive u. For example, consider the cardioid r = a(1 + cos u), shown in Fig. 13–18. Because dr>du = -a sin u, then when u = 30, tan c = a(1 + cos 30)>(-a sin 30) = -3.732, or c = -75, measured clockwise, opposite to +u as shown in the figure.

c

rc u

O

(c)

Fig. 13–17

Tangent

u r

procedure for analysis

c  75 r

Cylindrical or polar coordinates are a suitable choice for the analysis of a problem for which data regarding the angular motion of the radial line r are given, or in cases where the path can be conveniently expressed in terms of these coordinates. Once these coordinates have been established, the equations of motion can then be applied in order to relate the forces acting on the particle to its acceleration components. The method for doing this has been outlined in the procedure for analysis given in Sec. 13.4. The following is a summary of this procedure. Free-Body Diagram.

• Establish the r, u, z inertial coordinate system and draw the

particle’s free-body diagram. • Assume that ar , au , az act in the positive directions of r, u, z if they are unknown. • Identify all the unknowns in the problem. Equations of Motion. • Apply the equations of motion, Eq. 13–8. Kinematics. of Sec. 12.8 to determine r and the time • Use the methods # $ # $ $ derivatives r, r, u, u , #z, and then evaluate the acceleration $ # # $ $ components ar = r - ru2, au = ru + 2ru, az = z. • If any of the acceleration components is computed as a negative quantity, it indicates that it acts in its negative coordinate direction. • When taking the time derivatives of r = f(u), it is very important to use the chain rule of calculus, which is discussed in Appendix C.

13

ds

O

u  30 2a

Fig. 13–18

154

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13.10

ExaMplE

The smooth 0.5-kg double-collar in Fig. 13–19a can freely slide on arm AB and the circular # guide rod. If the arm rotates with a constant angular velocity of u = 3 rad>s, determine the force the arm exerts on the collar at the instant u = 45. Motion is in the horizontal plane.

B

C

13

Solution Free-body Diagram. The normal reaction NC of the circular guide rod and the force F of arm AB act on the collar in the plane of motion, Fig. 13–19b. Note that F acts perpendicular to the axis of arm AB, that is, in the direction of the u axis, while NC acts perpendicular to the tangent of the circular path at u = 45. The four unknowns are NC, F, ar, au.

r  (0.8 cos u) m u A u  3 rad/s

0.4 m

equations of Motion.

(a)

u

partiCle: forCe

NC

au

45 C

r ar tangent

+ QFr = mar:

-NC cos 45 = (0.5 kg) ar

(1)

+ aFu = mau:

F - NC sin 45 = (0.5 kg) au

(2)

Kinematics. Using the chain rule (see # Appendix $C), the first and second time derivatives of r when u = 45, u = 3 rad>s, u = 0, are r = 0.8 cos u = 0.8 cos 45 = 0.5657 m # # r = -0.8 sin u u = -0.8 sin 45(3) = -1.6971 m>s $ # $ r = -0.8 3 sin u u + cos u u2 4

F (b)

Fig. 13–19

= -0.8[sin 45(0) + cos 45(32)] = -5.091 m>s2

We have # $ ar = r - ru2 = -5.091 m>s2 - (0.5657 m)(3 rad>s)2 = -10.18 m>s2 $ # # au = ru + 2ru = (0.5657 m)(0) + 2(-1.6971 m>s)(3 rad>s) = -10.18 m>s2 Substituting these results into Eqs. (1) and (2) and solving, we get NC = 7.20 N F = 0

Ans.

13.6

155

equations of Motion: CylindriCal Coordinates

13.11

ExaMplE

The smooth 2-kg cylinder C in Fig. 13–20a has a pin P through its center which passes through the slot in arm OA. # If the arm is forced to rotate in the vertical plane at a constant rate u = 0.5 rad>s, determine the force that the arm exerts on the peg at the instant u = 60.

13

Solution Why is it a good idea to use polar coordinates to solve this problem? Free-body Diagram. The free-body diagram for the cylinder is shown in Fig. 13–20b. The force on the peg, FP , acts perpendicular to the slot in the arm. As usual, ar and au are assumed to act in the directions of positive r and u, respectively. Identify the four unknowns. equations of Motion. Using the data in Fig. 13–20b, we have + bFr = mar; + RFu = mau; Kinematics.

19.62 sin u - NC sin u = 2ar 19.62 cos u + FP - NC cos u = 2au

· u  0.5 rad/s 0.4 m r C

(1) (2)

From Fig. 13–20a, r can be related to u by the equation r =

O u

A (a)

0.4 = 0.4 csc u sin u

Since d(csc u) = -(csc u cot u) du and d(cot u) = -(csc2 u) du, then r and the necessary time derivatives become # u = 0.5 r = 0.4 csc u $ # # u = 0 r = -0.4(csc u cot u)u = -0.2 csc u cot u # # $ r = -0.2(-csc u cot u)(u) cot u - 0.2 csc u(-csc2 u)u = 0.1 csc u(cot2 u + csc2 u)

r # r $ r ar au

ar

NC

(b)

Fig. 13–20

= -0.133 0.192 # $ r - ru2 = 0.192 - 0.462(0.5)2 = 0.0770 $ # # ru + 2ru = 0 + 2(-0.133)(0.5) = -0.133

Substituting these results into Eqs. 1 and 2 with u = 60 and solving yields NC = 19.4 N

u

r

= 0.462 = = =

19.62 N FP

u

Evaluating these formulas at u = 60, we get # u = 0.5 $ u = 0

P

FP = -0.356 N

Ans.

The negative sign indicates that FP acts opposite to the direction shown in Fig. 13–20b.

au u

156

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and

a C C e l e r at i o n

13.12

r  0.1 u

13 A

KinetiCs

u

r

C

O · u  4 rad/s Top View

A can C, having a mass of 0.5 kg, moves along a grooved horizontal slot shown in Fig. 13–21a. The slot is in the form of a spiral, which is defined by the equation r = (0.1u)# m, where u is in radians. If the arm OA rotates with a constant rate u = 4 rad>s in the horizontal plane, determine the force it exerts on the can at the instant u = p rad. Neglect friction and the size of the can. Solution Free-body Diagram. The driving force FC acts perpendicular to the arm OA, whereas the normal force of the wall of the slot on the can, NC , acts perpendicular to the tangent to the curve at u = p rad, Fig. 13–21b. As usual, ar and au are assumed to act in the positive directions of r and u, respectively. Since the path is specified, the angle c which the extended radial line r makes with the tangent, Fig. 13–21c, can be determined from Eq. 13–9. We have r = 0.1u, so that dr>du = 0.1, and therefore

(a)

FC

tan c =

ar r

f NC f au

Tangent

u (b)

r 0.1u = = u dr>du 0.1

When u = p, c = tan-1p = 72.3, so that f = 90 - c = 17.7, as shown in Fig. 13–21c. Identify the four unknowns in Fig. 13–21b. equations of Motion. Fig. 13–21b, we have

Using f = 17.7 and the data shown in

+ Fr = mar; S

NC cos 17.7 = 0.5ar

(1)

+ T Fu = mau;

FC - NC sin 17.7 = 0.5au

(2)

Kinematics. The time derivatives of r and u are # u = 4 rad>s r = 0.1u $ # # u = 0 r = 0.1u = 0.1(4) = 0.4 m>s $ $ r = 0.1u = 0

r  0.1 u r c f Tangent

At the instant u = p rad, # $ ar = r - ru2 = 0 - 0.1(p)(4)2 = -5.03 m>s2 $ # # au = ru + 2ru = 0 + 2(0.4)(4) = 3.20 m>s2 Substituting these results into Eqs. 1 and 2 and solving yields

up

NC = -2.64 N

u (c)

Fig. 13–21

FC = 0.800 N What does the negative sign for NC indicate?

Ans.

13.6

157

equations of Motion: CylindriCal Coordinates

FunDaMental prObleMs # F13–13. Determine the constant angular velocity u of the vertical shaft of the amusement ride if f = 45. Neglect the mass of the cables and the size of the passengers.

1.5 m

F13–15. The 2-Mg car is traveling along the curved road described by r = (50e2u) m, where u is in radians. If a camera is located at A and it rotates with an angular # velocity of u = 0.05 rad>s and an angular acceleration of 13 $ u = 0.01 rad>s2 at the instant u = p6 rad, determine the resultant friction force developed between the tires and the road at this instant.

u 8m

r  (50e2u) m

f

r

Prob. F13–13

u A u, u

Prob. F13–15

F13–14. The 0.2-kg ball is blown through the smooth vertical circular tube whose shape is defined by r = (0.6 sin u) m, where u is in radians. If u = (p t2) rad, where t is in seconds, determine the magnitude of force F exerted by the blower on the ball when t = 0.5 s.

F13–16. The 0.2-kg pin P is constrained to move in the smooth curved slot, which is defined by the lemniscate r = (0.6 cos 2u) m. Its motion is controlled by the rotation of the slotted arm # OA, which has a constant clockwise angular velocity of u = -3 rad>s. Determine the force arm OA exerts on the pin P when u = 0. Motion is in the vertical plane.

F

r  (0.6 cos 2u) m

0.3 m r u

Prob. F13–14

u O

u

Prob. F13–16

P

A

158

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prObleMs 13–85. Using air pressure, the 0.5-kg ball is forced to move through the tube lying in the horizontal plane and having 13 the shape of a logarithmic spiral. If the tangential force exerted on the ball due to the air is 6 N, determine the rate of increase in the ball’s speed at the instant u = p>2. What direction does it act in? 13–86. Solve Prob. 13–85 if the tube lies in a vertical plane.

*13–88. Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 u) m, where u is in radians. If the angular position of the arm is u = (p8 t2) rad, where t is in seconds, determine the force of the rod on the peg and the normal force of the slot on the peg at the instant t = 2 s. The peg is in contact with only one edge of the rod and slot at any instant.

0.2e0.1u

r

P r  (0.5) m

r

r

 F

6N

u

Prob. 13–88 Probs. 13–85/86

13–87. The 0.75-kg smooth can is guided along the circular path using the arm guide. If the arm has an$ angular velocity # u = 2 rad>s and an angular acceleration u = 0.4 rad>s2 at the instant u = 30, determine the force of the guide on the can. Motion occurs in the horizontal plane.

r

# 13–89. The arm is rotating at a rate of u = 4 rad>s when $ u = 3 rad>s2 and u = 180. Determine the force it must exert on the 0.5-kg smooth cylinder if it is confined to move along the slotted path. Motion occurs in the horizontal plane.

r

·

··   4 rad/s,   3 rad/s2

u   180

0.5 m

2 r  (—) m 

0.5 m

Prob. 13–87

Prob. 13–89

13.6 13–90. The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, u = (0.7t) rad, and z = (- 0.5t) m, where t is in seconds. Determine the components of force Fr, Fu, and Fz which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy.

159

equations of Motion: CylindriCal Coordinates

*13–92. The tube # rotates in the horizontal plane at a constant rate of u = 4 rad>s. If a 0.2-kg ball # B starts at the origin O with an initial radial velocity r = 1.5 m>s and moves outward through the tube, determine the radial and transverse components of the ball’s velocity at the instant it leaves the outer end at C, r = 0.5 m. Hint: Show that the $ equation of motion in the r direction is r - 16r = 0. 13 - 4t 4t The solution is of the form r = Ae + Be . Evaluate the integration constants A and B, and determine the time t when r = 0.5 m. Proceed to obtain vr and vu.

z u

z

r  1.5 m

· u

4 rad/s

O u

z

r

13–91. The 40-kg boy is sliding down the smooth spiral slide such that z = - 2 m>s and his speed is 2 m>s. Determine the r, u, z components of force the slide exerts on him at this instant. Neglect the size of the boy.

C

Prob. 13–92

13–93. The girl has a mass of 50 kg. She is seated on the horse of the merry-go-round which undergoes constant # rotational motion u = 1.5 rad>s. If the path of the horse is defined by r = 4 m, z = (0.5 sin u) m, determine the maximum and minimum force Fz the horse exerts on her during the motion.

z u

z

r  1.5 m

u r

z

Prob. 13–91

y

B

x

Prob. 13–90

0.5 m

Prob. 13–93

160

Chapter 13

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partiCle: forCe

13–94. Using a forked rod, a smooth cylinder P, having a mass of 0.4 kg, is forced to move along the vertical slotted path r = (0.6u) m, where u is in radians. If the cylinder has a constant speed of vC = 2 m>s, determine the force of the rod and the normal force of the slot on the cylinder at the instant u = p rad. Assume the cylinder is in contact with only one 13 edge of the rod and slot at any instant. Hint: To obtain the time derivatives necessary to compute the cylinder’s acceleration components ar and au, take the first and second time derivatives of r = 0.6u.# Then, for further information, use Eq. 12–26 to determine u. Also, take the time derivative $ # of Eq. 12–26, noting that v = 0 to determine u .

and

a C C e l e r at i o n

*13–96. The particle has a mass of 0.5 kg and is confined to move along the smooth vertical slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when u. = 30. The rod is rotating with a constant angular velocity u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant.

A P

r

up

r

u O

r  0.6u



u = 2 rad/s

Prob. 13–94 0.5 m

13–95. A car of a roller coaster travels along a track which for a short distance is defined by a conical spiral, r = 34z, u = - 1.5z, where r and# z are in meters and u in radians. If the angular motion u = 1 rad>s is always maintained, determine the r, u, z components of reaction exerted on the car by the track at the instant z = 6 m. The car and passengers have a total mass of 200 kg.

Prob. 13–96

13–97. A smooth can C, having a mass of 3 kg, is lifted from a feed at A to a ramp at B by a rotating rod. If the rod # maintains a constant angular velocity of u = 0.5 rad>s, determine the force which the rod exerts on the can at the instant u = 30. Neglect the effects of friction in the calculation and the size of the can so that r = (1.2 cos u) m. The ramp from A to B is circular, having a radius of 600 mm.

B

z · u  0.5 rad/s r

u

Prob. 13–95

C r 600 mm

u 600 mm

Prob. 13–97

A

13.6 13–98. The spring-held follower AB has a mass of 0.5 kg and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.15 m and z = (0.02 cos 2u) m. If the cam is rotating at a constant rate of 30 rad>s, determine the force component Fz at the end A of the follower when u = 30. The spring is uncompressed when u = 90. Neglect friction at the bearing C.

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*13–100. Determine the normal and frictional driving forces that the partial spiral track exerts on# the 200-kg motorcycle at the instant u = 53p rad, u = 0.4 rad>s, $ u = 0.8 rad>s2. Neglect the size of the motorcycle. 13

r  (5u) m u r

z  (0.02 cos 2) m z 0.15 m ·

C B

  30 rad/s

A

k  1000 N/m

Prob. 13–100 Prob. 13–98

13–99. The spring-held follower AB has a mass of 0.5 kg and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.15 m and z = (0.02 cos 2u) m. If the cam is rotating at a constant rate of 30 rad>s, determine the maximum and minimum force components Fz the follower exerts on the cam if the spring is uncompressed when u = 90.

13–101. The 0.5-kg ball is guided along the vertical circular path r = 2rc cos. u using the arm OA. If the arm has an angular velocity u = 0.4 rad>s and an angular acceleration $ u = 0.8 rad>s2 at the instant u = 30, determine the force of the arm on the ball. Neglect friction and the size of the ball. Set rc = 0.4 m. 13–102. The ball of mass m is guided along the vertical circular path r = 2rc cos u using the arm OA. If the arm has . a constant angular velocity u 0, determine the angle u … 45 at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball.

P z  (0.02 cos 2) m z 0.15 m ·

  30 rad/s

r

C B A

k  1000 N/m

rc

u O

Prob. 13–99

Probs. 13–101/102

A

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13–103. Rod . OA rotates counterclockwise at a constant angular rate u = 4 rad>s. The double collar B is pin-connected together such that one collar slides over the rotating rod and the other collar slides over the circular rod described by the equation r = (1.6 cos u) m. If both collars have a mass of 0.5 kg, determine the force which the circular rod exerts on one 13 of the collars and the force that OA exerts on the other collar at the instant u = 45. Motion is in the horizontal plane.

and

a C C e l e r at i o n

13–106. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = (eu) m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u. = 45, if the force F maintains a constant angular motion u = 2 rad>s.

*13–104. Solve Prob. 13–103 if motion is in the vertical plane. A F

r  1.6 cos u •

u = 4 rad/s

B u r  eθ

r

O u

0.8 m

Prob. 13–106 Probs. 13–103/104 13–105. The smooth surface of the vertical cam is defined in part by the curve r = (0.2 cos u + 0.3) m. $ The forked rod is rotating with an angular acceleration of u = 2. rad>s2, and when u = 45 the angular velocity is u = 6 rad>s. Determine the force the cam and the rod exert on the 2-kg roller at this instant. The attached spring has a stiffness k = 100 N/m and an unstretched length of 0.1 m.

13–107. The pilot of the airplane executes a vertical loop which in part follows the path of a cardioid, r = 200(1 + cos u) m, where u is in radians. If his speed at A is a constant vp = 85 m>s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at A. He has a mass of 80 kg. Hint: To determine the time derivatives necessary to calculate the acceleration components ar and au, take the first and second time derivatives of r = 200(1 + cos u). Then, for further # information, use Eq. 12–26 to determine u.

r

u

 r  200 (1  cos  ) m

A

Prob. 13–105

Prob. 13–107

13.6 *13–108. The pilot of an airplane executes a vertical loop which in part follows the path of a “four-leaved rose,” r = ( - 180 cos 2u) m, where u is in radians. If his speed at A is a constant vp = 24 m>s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at A. He weights 650 N. Hint: To determine the time derivatives necessary to compute the acceleration components ar and  au, take the first and second time derivatives of for further information, use r = -180(1 + cos u). Then, # Eq. 12–26 to determine u. Also, take the time # derivative of # Eq. 12–26, noting that yC = 0 to determine u.

equations of Motion: CylindriCal Coordinates

163

13–111. A 0.2-kg spool slides down along a smooth rod. If# the rod has a constant angular rate of rotation u = 2 rad>s in the vertical plane, show that the equations $ of motion for the spool are r - 4r - 9.81 sin u = 0 # and 0.8r + N s - 1.962 cos u = 0, where N s is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that 13 the solution of the first of these equations is # r = C1 e-2t + C2 e2t - (9.81>8) sin 2t. If r, r, and u are zero when t = 0, evaluate the constants C1 and C2 to determine r at the instant u = p>4 rad.

A 24 m/s

u r 180 cos 2u

r

u  2 rad/s

u

r

Prob. 13–111

Prob. 13–108 13–109. The particle has a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when u =. 30. The rod is rotating with a constant angular velocity u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant. 13–110. Solve $Prob. 13–109 if the # arm has an angular acceleration of u = 3 rad>s2 when u = 2 rad>s at u = 30.

*13–112. The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius r0 = 0.5 m such that the angular rate of rotation is . u0 = 1 rad>s. If the attached cord ABC is drawn down through the hole at a constant speed of 0.2 m>s, determine the tension the cord exerts on the ball at the instant r = 0.25 m. Also, compute the angular velocity of the ball at this instant. Neglect the effects of friction between the ball and horizontal plane. Hint: First show$ that the equation of# motion in the u # # direction yields au = ru + 2ru = (1/r)(d(r2u)/dt) = 0. When # integrated, r2u = c, where the constant c is determined from the problem data.

A

A · u  2 rad/s

r B

r

u u0

0.5 m

r0

u

0.2 m/s C

O

F

Probs. 13–109/110

Prob. 13–112

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*13.7 P

13 F du

r u

1 dA  — r 2du 2

O

If a particle is moving only under the influence of a force having a line of action which is always directed toward a fixed point, the motion is called central-force motion. This type of motion is commonly caused by electrostatic and gravitational forces. In order to analyze the motion, we will consider the particle P shown in Fig. 13–22a, which has a mass m and is acted upon only by the central force F. The free-body diagram for the particle is shown in Fig. 13–22b. Using polar coordinates (r, u), the equations of motion, Eq. 13–8, become

-F = m c

d2r du 2 r a b d dt dt2

Fu = mau; u r

(13–10)

0 = mar

d2u dr du + 2 b 2 dt dt dt

The second of these equations may be written in the form

u

F

a C C e l e r at i o n

Central-Force Motion and Space Mechanics

Fr = mar; (a)

and

so that integrating yields

1 d 2 du c ar bd = 0 r dt dt

(b)

r2

Fig. 13–22

du = h dt

(13–11)

Here h is the constant of integration. From Fig. 13–22a notice that the shaded area described by the radius r, as r moves through an angle du, is dA = 12 r2 du. If the areal velocity is defined as dA 1 du h = r2 = (13–12) dt 2 dt 2 then it is seen that the areal velocity for a particle subjected to centralforce motion is constant. In other words, the particle will sweep out equal segments of area per unit of time as it travels along the path. To obtain the path of motion, r = f(u), the independent variable t must be eliminated from Eq. 13–10. Using the chain rule of calculus and Eq. 13–11, the time derivatives of Eq. 13–10 may be replaced by dr dr du h dr = = 2 dt du dt r du d2r d h dr d h dr du d h dr h = a 2 b = a 2 b = c a 2 bd 2 2 dt du du du dt du du dt r r r r

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Substituting a new dependent variable (xi) j = 1>r into the second equation, we have d2j d2r = -h2j2 2 2 dt du 13

Also, the square of Eq. 13–11 becomes a

du 2 b = h2j4 dt

Substituting these two equations into the first of Eq. 13–10 yields -h2j2 or

d2j F - h2j3 = m du2

d2j F + j = 2 du mh2j2

This satellite is subjected to a central force and its orbital motion can be closely predicted using the equations developed in this section. (UniversalImagesGroup/ Getty Images)

(13–13)

This differential equation defines the path over which the particle travels when it is subjected to the central force F.* For application, the force of gravitational attraction will be considered. Some common examples of central-force systems which depend on gravitation include the motion of the moon and artificial satellites about the earth, and the motion of the planets about the sun. As a typical problem in space mechanics, consider the trajectory of a space satellite or space vehicle launched into free-flight orbit with an initial velocity v0 , Fig. 13–23. It will be assumed that this velocity is initially parallel to the tangent at the surface of the earth, as shown in the figure.† Just after the satellite is released into free flight, the only force acting on it is the gravitational force of the earth. (Gravitational attractions involving other bodies such as the moon or sun will be neglected, since for orbits close to the earth their effect is small in comparison with the earth’s gravitation.) According to Newton’s law of gravitation, force F will always act between the mass centers of the earth and the satellite, Fig. 13–23. From Eq. 13–1, this force of attraction has a magnitude of F = G

Mem r2

where Me and m represent the mass of the earth and the satellite, respectively, G is the gravitational constant, and r is the distance between *In the derivation, F is considered positive when it is directed toward point O. If F is oppositely directed, the right side of Eq. 13–13 should be negative. †The case where v acts at some initial angle u to the tangent is best described using the 0 conservation of angular momentum.

Free-flight trajectory v0 Satellite F Power-flight trajectory F

r  r0

Launching

Fig. 13–23

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the mass centers. To obtain the orbital path, we set j = 1>r in the foregoing equation and substitute the result into Eq. 13–13. We obtain GMe d2j + j = 2 du h2 13

(13–14)

This second-order differential equation has constant coefficients and is nonhomogeneous. The solution is the sum of the complementary and particular solutions given by j = D directrix P

A

uf r

x

focus F

f u

GMe 1 = C cos (u - f) + r h2

This equation represents the free-flight trajectory of the satellite. It is the equation of a conic section expressed in terms of polar coordinates. A geometric interpretation of Eq. 13–15 requires knowledge of the equation for a conic section. As shown in Fig. 13–24, a conic section is defined as the locus of a point P that moves in such a way that the ratio of its distance to a focus, or fixed point F, to its perpendicular distance to a fixed line DD called the directrix, is constant. This constant ratio will be denoted as e and is called the eccentricity. By definition FP PA

e = x¿

p D

(13–15)

From Fig. 13–24,

Fig. 13–24

FP = r = e(PA) = e[p - r cos(u - f)] or 1 1 1 = cos(u - f) + r p ep Comparing this equation with Eq. 13–15, it is seen that the fixed distance from the focus to the directrix is

p =

1 C

(13–16)

And the eccentricity of the conic section for the trajectory is

e =

Ch2 GMe

(13–17)

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167

Provided the polar angle u is measured from the x axis (an axis of symmetry since it is perpendicular to the directrix), the angle f is zero, Fig. 13–24, and therefore Eq. 13–15 reduces to GMe 1 = C cos u + r h2

(13–18)

The constants h and C are determined from the data obtained for the position and velocity of the satellite at the end of the power-flight trajectory. For example, if the initial height or distance to the space vehicle is r0, measured from the center of the earth, and its initial speed is v0 at the beginning of its free flight, Fig. 13–25, then the constant h may be obtained from Eq. 13–11. When u = f = 0, the velocity v0 has no radial component; therefore, from Eq. 12–25, v0 = r0(du>dt), so that h = r20

du dt

or h = r0v0

(13–19)

To determine C, use Eq. 13–18 with u = 0, r = r0 , and substitute Eq. 13–19 for h: C =

GMe 1 a1 b r0 r0v20

(13–20)

The equation for the free-flight trajectory therefore becomes GMe GMe 1 1 = a1 b cos u + 2 2 2 r r0 r0v0 r0v0

(13–21)

The type of path traveled by the satellite is determined from the value of the eccentricity of the conic section as given by Eq. 13–17. If e e e e

= = 6 7

0 1 1 1

[email protected] trajectory is a circle [email protected] trajectory is a parabola [email protected] trajectory is an ellipse [email protected] trajectory is a hyperbola

(13–22)

13

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and

Parabolic Path. Each of these trajectories is shown in Fig. 13–25. From the curves it is seen that when the satellite follows a parabolic path, it is “on the border” of never returning to its initial starting point. The initial launch velocity, v0 , required for the satellite to follow a parabolic path is called the escape velocity. The speed, ve , can be determined by using the second of Eq. 13–22, e = 1, with Eqs. 13–17, 13–19, and 13–20. It is left as an exercise to show that ve =

2GMe B r0

(13–23)

Circular Orbit. The speed vc required to launch a satellite into a circular orbit can be found using the first of Eq. 13–22, e = 0. Since e is related to h and C, Eq. 13–17, C must be zero to satisfy this equation (from Eq. 13–19, h cannot be zero); and therefore, using Eq. 13–20, we have vc =

GMe B r0

(13–24)

Provided r0 represents a minimum height for launching, in which frictional resistance from the atmosphere is neglected, speeds at launch which are less than vc will cause the satellite to reenter the earth’s atmosphere and either burn up or crash, Fig. 13–25.

Hyperbolic trajectory e1 Parabolic trajectory e1 Elliptical trajectory e1 v0

Crash trajectory v0  vc

r0

Fig. 13–25

Circular trajectory e0

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169

b rp O

ra

13 b

a

a

Fig. 13–26

Elliptical Orbit. All the trajectories attained by planets and most

satellites are elliptical, Fig. 13–26. For a satellite’s orbit about the earth, the minimum distance from the orbit to the center of the earth O (which is located at one of the foci of the ellipse) is rp and can be found using Eq. 13–21 with u = 0. Therefore; rp = r 0

(13–25)

This minimum distance is called the perigee of the orbit. The apogee or maximum distance ra can be found using Eq. 13–21 with u = 180.* Thus, ra =

r0 (2GMe >r0y20) - 1

(13–26)

With reference to Fig. 13–26, the half-length of the major axis of the ellipse is a =

r p + ra 2

(13–27)

Using analytical geometry, it can be shown that the half-length of the minor axis is determined from the equation b = 2rpra

(13–28)

*Actually, the terminology perigee and apogee pertains only to orbits about the earth. If any other heavenly body is located at the focus of an elliptical orbit, the minimum and maximum distances are referred to respectively as the periapsis and apoapsis of the orbit.

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Furthermore, by direct integration, the area of an ellipse is A = pab = 13

p (r + ra)2rpra 2 p

(13–29)

The areal velocity has been defined by Eq. 13–12, dA>dt = h>2. Integrating yields A = hT>2, where T is the period of time required to make one orbital revolution. From Eq. 13–29, the period is

T =

p (r + ra)2rpra h p

(13–30)

In addition to predicting the orbital trajectory of earth satellites, the theory developed in this section is valid, to a surprisingly close approximation, at predicting the actual motion of the planets traveling around the sun. In this case the mass of the sun, Ms , should be substituted for Me when the appropriate formulas are used. The fact that the planets do indeed follow elliptic orbits about the sun was discovered by the German astronomer Johannes Kepler in the early seventeenth century. His discovery was made before Newton had developed the laws of motion and the law of gravitation, and so at the time it provided important proof as to the validity of these laws. Kepler’s laws, developed after 20 years of planetary observation, are summarized as follows: 1. Every planet travels in its orbit such that the line joining it to the center of the sun sweeps over equal areas in equal intervals of time, whatever the line’s length.

2. The orbit of every planet is an ellipse with the sun placed at one of its foci.

3. The square of the period of any planet is directly proportional to the cube of the major axis of its orbit. A mathematical statement of the first and second laws is given by Eqs. 13–12 and 13–21, respectively. The third law can be shown from Eq. 13–30 using Eqs. 13–18, 13–27, and 13–28. (See Prob. 13–117.)

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prObleMs In the following problems, except where otherwise indicated, assume that the radius of the earth is 6378 km, the earth’s mass is 5.976(1024) kg, the mass of the sun is 1.99(1030) kg, and the gravitational constant is G = 66.73(10 - 12) m3 >(kg # s2).

13–113. The earth has an orbit with eccentricity 0.0167 around the sun. Knowing that the earth’s minimum distance from the sun is 146(106) km, find the speed at which the earth travels when it is at this distance. Determine the equation in polar coordinates which describes the earth’s orbit about the sun.

13–114. A communications satellite is in a circular orbit above the earth such that it always remains directly over a point on the earth’s surface. As a result, the period of the satellite must equal the rotation of the earth, which is approximately 24 hours. Determine the satellite’s altitude h above the earth’s surface and its orbital speed. 13–115. The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13–24. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit 800 km from the earth’s surface.

13–118. The satellite is moving in an elliptical orbit with an eccentricity e = 0.25. Determine its speed when it is at  its maximum distance A and minimum distance B from the earth. 13

2 Mm

A

B

Prob. 13–118 13–119. The rocket is traveling in free flight along the elliptical orbit. The planet has no atmosphere, and its mass is 0.60 times that of the earth. If the rocket has the orbit shown, determine the rocket’s speed when it is at A and at B.

B

O

*13–116. The rocket is in circular orbit about the earth at an altitude of 20 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field.

18.3 Mm

A

7.60 Mm

Prob. 13–119

20 Mm

*13–120. Determine the constant speed of satellite S so that it circles the earth with an orbit of radius r = 15 Mm. Hint: Use Eq. 13–1.

r  15 Mm

Prob. 13–116 S

13–117. Prove Kepler’s third law of motion. Hint: Use Eqs. 13–18, 13–27, 13–28, and 13–30.

Prob. 13–120

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13–121. The rocket is in free flight along an elliptical trajectory A A . The planet has no atmosphere, and its mass is 0.70 times that of the earth. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A. 13

and

a C C e l e r at i o n

13–123. The rocket is initially in free-flight circular orbit around the earth. Determine the speed of the rocket at A. What change in the speed at A is required so that it can move in an elliptical orbit to reach point A ? *13–124. The rocket is in free-flight circular orbit around the earth. Determine the time needed for the rocket to travel from the inner orbit at A to the outer orbit at A .

8 Mm

r  3 Mm A

B

O

A¿

A O

6 Mm

A¿ 19 Mm

9 Mm

Probs. 13–123/124

Prob. 13–121

13–122. The Viking Explorer approaches the planet Mars on a parabolic trajectory as shown. When it reaches point A its velocity is 10 Mm>h. Determine r0 and the required change in velocity at A so that it can then maintain a circular orbit as shown. The mass of Mars is 0.1074 times the mass of the earth.

A

13–125. A rocket is in a free-flight elliptical orbit about the earth such that the eccentricity of its orbit is e and its perigee is r0. Determine the minimum increment of speed it should have in order to escape the earth’s gravitational field when it is at this point along its orbit. 13–126. The rocket is traveling around the earth in free flight along the elliptical orbit. If the rocket has the orbit shown, determine the speed of the rocket when it is at A and at B.

B

r0

Prob. 13–122

A

30 Mm

20 Mm

Prob. 13–126

13.7 13–127. An elliptical path of a satellite has an eccentricity e = 0.130. If it has a speed of 15 Mm>h when it is at perigee, P, determine its speed when it arrives at apogee, A. Also, how far is it from the earth’s surface when it is at A?

Central-forCe Motion and spaCe MeChaniCs

173

13–129. The rocket is traveling in a free flight along an elliptical trajectory A A. The planet has no atmosphere, and its mass is 0.60 times that of the earth. If the rocket has the orbit shown, determine the rocket’s velocity when it is at point A. 13–130. If the rocket is to land on the surface of the planet, 13 determine the required free-flight speed it must have at A  so that the landing occurs at B. How long does it take for the rocket to land, going from A  to B? The planet has no atmosphere, and its mass is 0.6 times that of the earth.

A

P

r  6 Mm A¿

O

B

A

Prob. 13–127

100 Mm

*13–128. A rocket is in free-flight elliptical orbit around the planet Venus. Knowing that the periapsis and apoapsis of the orbit are 8 Mm and 26 Mm, respectively, determine (a) the speed of the rocket at point A , (b) the required speed it must attain at A just after braking so that it undergoes an 8-Mm free-flight circular orbit around Venus, and (c) the periods of both the circular and elliptical orbits. The mass of Venus is 0.816 times the mass of the earth.

O

A¿

A

70 Mm

Probs. 13–129/130 13–131. The rocket is traveling around the earth in free flight along an elliptical orbit AC. If the rocket has the orbit shown, determine the rocket’s velocity when it is at point A. *13–132. The rocket is traveling around the earth in free flight along the elliptical orbit AC. Determine its change in speed when it reaches A so that it travels along the elliptical orbit AB.

C

A

B

8 Mm

8 Mm

18 Mm

Prob. 13–128

8 Mm

Probs. 13–131/132

10 Mm

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13

C13–1. If the box is released from rest at A, use numerical values to show how you would estimate the time for it to arrive at B. Also, list the assumptions for your analysis.

C13–3. Determine the smallest speed of each car A and B so that the passengers do not lose contact with the seat while the arms turn at a constant rate. What is the largest normal force of the seat on each passenger? Use numerical values to explain your answer.

B A

A B

Prob. C13–1 C13–2. The tugboat has a known mass and its propeller provides a known maximum thrust. When the tug is fully powered you observe the time it takes for the tug to reach a speed of known value starting from rest. Show how you could determine the mass of the barge. Neglect the drag force of the water on the tug. Use numerical values to explain your answer.

Prob. C13–3 C13–4. Each car is pin connected at its ends to the rim of the wheel which turns at a constant speed. Using numerical values, show how to determine the resultant force the seat exerts on the passenger located in the top car A. The passengers are seated toward the center of the wheel. Also, list the assumptions for your analysis.

A

Prob. C13–2 Prob. C13–4

175

Chapter review

Chapter review Kinetics Kinetics is the study of the relation between forces and the acceleration they cause. This relation is based on Newton’s second law of motion, expressed mathematically as F = ma.

F2 FR = F

Before applying the equation of motion, it is important to first draw the particle’s free-body diagram in order to account for all of the forces that act on the particle. Graphically, this diagram is equal to the kinetic diagram, which shows the result of the forces, that is, the ma vector.

F1 Free-body diagram

Rectangular x, y, z axes are used to describe the motion along each of the axes.

Normal, tangential, and binormal axes n, t, b, are often used when the path is known. Recall that an is always directed in the + n direction. It indicates the change in the velocity direction. Also recall that at is tangent to the path. It indicates the change in the velocity magnitude.

Cylindrical coordinates are useful when angular motion of the radial line r is specified or when the path can conveniently be described with these coordinates.

Kinetic diagram

y

Inertial Coordinate Systems When applying the equation of motion, it is important to measure the acceleration from an inertial coordinate system. This system has axes that do not rotate but are either fixed or translate with a constant velocity. Various types of inertial coordinate systems can be used to apply F = ma in component form.

13

ma



a Path of particle vO O

Inertial frame of reference

x

Fx = max , Fy = may , Fz = maz

Ft = mat , Fn = man , Fb = 0 at = dv>dt or at = v dv>ds an = v2 >r

where r =

[1 + (dy>dx)2]3>2  d2y>dx2 

# $ Fr = m(r - ru2) $ ## Fu = m(ru + 2ru) $ Fz = mz

Central-Force Motion When a single force acts upon a particle, such as during the free-flight trajectory of a satellite in a gravitational field, then the motion is referred to as central-force motion. The orbit depends upon the eccentricity e; and as a result, the trajectory can either be circular, parabolic, elliptical, or hyperbolic.

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review prObleMs R13–1. The van is traveling at 20 km>h when the coupling of the trailer at A fails. If the trailer has a mass of 250 kg and 13 coasts 45 m before coming to rest, determine the constant horizontal force F created by rolling friction which causes the trailer to stop.

R13–3. Block B rests on a smooth surface. If the coefficients of friction between A and B are ms = 0.4 and mk = 0.3, determine the acceleration of each block if F = 250 N.

20 km/h 10 kg F

A

A F

25 kg

B

Prob. R13–1 Prob. R13–3

R13–2. The motor M pulls in its attached rope with an acceleration ap = 6 m>s2. Determine the towing force exerted by M on the rope in order to move the 50-kg crate up the inclined plane. The coefficient of kinetic friction between the crate and the plane is mk = 0.3. Neglect the mass of the pulleys and rope.

R13–4. If the motor draws in the cable at a rate of v = (0.05s3>2) m>s, where s is in meters, determine the tension developed in the cable when s = 10 m. The crate has a mass of 20 kg, and the coefficient of kinetic friction between the crate and the ground is mk = 0.2.

aP  6 m/s2 M

P

s v M

u  30

Prob. R13–2

Prob. R13–4

177

review probleMs R13–5. The ball has a mass of 30 kg and a speed v = 4 m>s at the instant it is at its lowest point, u = 0. Determine the tension in the cord and the rate at which the ball’s speed is decreasing at the instant u = 20. Neglect the size of the ball.

R13–7. The 5-kg suitcase slides down the curved ramp for which the coefficient of kinetic friction is mk = 0.2. If at the instant it reaches point A it has a speed of 2 m>s, determine the normal force on the suitcase and the rate of increase of its speed. 13 y

1 x2 y  –– 8 3m u A

4m

x

Prob. R13–7

Prob. R13–5

R13–6. The bottle rests at a distance of 1.5 m from the center of the horizontal platform. If the coefficient of static friction between the bottle and the platform is ms = 0.3, determine the maximum speed that the bottle can attain before slipping. Assume the angular motion of the platform is slowly increasing.

R13–8. The spool, which has a mass of 4 kg, slides along the rotating rod. At the instant shown, the angular rate of # rotation of the rod is u = 6 rad>s and this rotation is $ increasing at u = 2 rad>s2. At this same instant, the spool has a velocity of 3 m>s and an acceleration of 1 m>s2, both measured relative to the rod and directed away from the center O when r = 0.5 m. Determine the radial frictional force and the normal force, both exerted by the rod on the spool at this instant.

r  0.5 m ••

u  2 rad/s2 O 1.5 m



u  6 rad/s

vs  3 m/s Motion

Prob. R13–6

as  1 m/s2

Prob. R13–8

Chapter 14

(© Oliver Furrer/Ocean/Corbis) As the woman falls, her energy will have to be absorbed by the bungee cord. The principles of work and energy can be used to predict the motion.

Kinetics of a Particle: Work and Energy Video Solutions are available for selected questions in this chapter.

Chapter ObjeCtives n

To develop the principle of work and energy and apply it to solve problems that involve force, velocity, and displacement.

n

To study problems that involve power and efficiency.

n

To introduce the concept of a conservative force and apply the theorem of conservation of energy to solve kinetic problems.

14.1

The Work of a Force

In this chapter, we will analyze motion of a particle using the concepts of work and energy. The resulting equation will be useful for solving problems that involve force, velocity, and displacement. Before we do this, however, we must first define the work of a force. Specifically, a force F will do work on a particle only when the particle undergoes a displacement in the direction of the force. For example, if the force F in Fig. 14–1 causes the particle to move along the path s from position r to a new position r, the displacement is then dr = r - r. The magnitude of dr is ds, the length of the differential segment along the path. If the angle between the tails of dr and F is u, Fig. 14–1, then the work done by F is a scalar quantity, defined by dU = F ds cos u

F

u

ds

dr r¿

r s

Fig. 14–1

180

Chapter 14

KinetiCs

partiCle: WorK

and

energy

By definition of the dot product (see Eq. B–14) this equation can also be written as

F

u

of a

dU = F # dr

ds

dr r¿

r s

14

Fig. 14–1 (Repeated)

This result may be interpreted in one of two ways: either as the product of F and the component of displacement ds cos u in the direction of the force, or as the product of ds and the component of force, F cos u, in the direction of displacement. Note that if 0 … u 6 90, then the force component and the displacement have the same sense so that the work is positive; whereas if 90 6 u … 180, these vectors will have opposite sense, and therefore the work is negative. Also, dU = 0 if the force is perpendicular to displacement, since cos 90 = 0, or if the force is applied at a fixed point, in which case the displacement is zero. The unit of work in SI units is the joule (J), which is the amount of work done by a one-newton force when it moves through a distance of one meter in the direction of the force (1 J = 1 N # m).

Work of a Variable Force. If the particle acted upon by the

force F undergoes a finite displacement along its path from r1 to r2 or s1 to s2 , Fig. 14–2a, the work of force F is determined by integration. Provided F and u can be expressed as a function of position, then r2

U1 - 2 =

Lr1

F # dr =

s2

Ls1

(14–1)

F cos u ds

Sometimes, this relation may be obtained by using experimental data to plot a graph of F cos u vs. s. Then the area under this graph bounded by s1 and s2 represents the total work, Fig. 14–2b.

F

F cos u

s2

F cos u

u F cos u s1

r2 s1

r1

s

ds

(b)

(a)

Fig. 14–2

s2

s

181

the WorK of a forCe

14.1 F cos u Fc

Fc cos u

u

s

s1

Fc cos u

s2

(a)

s

s2

s1 (b)

Fig. 14–3

14

Work of a Constant Force Moving Along a Straight Line.

If the force Fc has a constant magnitude and acts at a constant angle u from its straight-line path, Fig. 14–3a, then the component of Fc in the direction of displacement is always Fc cos u. The work done by Fc when the particle is displaced from s1 to s2 is determined from Eq. 14–1, in which case s2

U1 - 2 = Fc cos u or

Ls1

ds

U1 - 2 = Fc cos u(s2 - s1)

(14–2)

The crane must do work in order to hoist the weight of the pipe.

Here the work of Fc represents the area of the rectangle in Fig. 14–3b.

Work of a Weight. Consider a particle of weight W, which moves up along the path s shown in Fig. 14–4 from position s1 to position s2 . At an intermediate point, the displacement dr = dxi + dyj + dzk. Since W = -Wj, applying Eq. 14–1 we have U1 - 2 =

L

F # dr =

r2

Lr1

(-Wj) # (dxi + dyj + dzk)

y

y2

= or

Ly1

-W dy = -W(y2 - y1)

U1 - 2 = -W y

W

s2

dr

(14–3)

Thus, the work is independent of the path and is equal to the magnitude of the particle’s weight times its vertical displacement. In the case shown in Fig. 14–4 the work is negative, since W is downward and y is upward. Note, however, that if the particle is displaced downward (- y), the work of the weight is positive. Why?

s1 r1

s

r2 y2

y1

z

Fig. 14–4

x

182

Chapter 14

KinetiCs

of a

partiCle: WorK

and

energy

Work of a Spring Force.

If an elastic spring is elongated a distance ds, Fig. 14–5a, then the work done by the force that acts on the attached particle is dU = -Fsds = -ks ds. The work is negative since Fs acts in the opposite sense to ds. If the particle displaces from s1 to s2, the work of Fs is then s2

U1 - 2 =

Ls1

s2

Fs ds =

Ls1

-ks ds

U1 - 2 = - 1 12 ks22 - 12 ks21 2

14

(14–4)

This work represents the trapezoidal area under the line Fs = ks, Fig. 14–5b. A mistake in sign can be avoided when applying this equation if one simply notes the direction of the spring force acting on the particle and compares it with the sense of direction of displacement of the particle— if both are in the same sense, positive work results; if they are opposite to one another, the work is negative. Unstretched position, s  0 Fs Fs  ks ds s Fs

k

Force on Particle

s1

s

s2 (b)

(a)

Fig. 14–5 T f

u

The forces acting on the cart, as it is pulled a distance s up the incline, are shown on its free-body diagram. The constant towing force T does positive work of UT = (T cos f)s, the weight does negative work of UW = -(W sin u)s, and the normal force N does no work since there is no displacement of this force along its line of action.

W

u

N

14.1

ExamplE

the WorK of a forCe

183

14.1

The 10-kg block shown in Fig. 14–6a rests on the smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all the forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m.

s2m

Initial position of spring

P  400 N

2 sin 30 m

k  30 N/m

Solution First the free-body diagram of the block is drawn in order to account for all the forces that act on the block, Fig. 14–6b. horizontal Force P. Since this force is constant, the work is determined using Eq. 14–2. The result can be calculated as the force times the component of displacement in the direction of the force; i.e.,

2 cos 30 m

98.1 N 30 P  400 N 30

UP = 400 N (2 m cos 30) = 692.8 J or the displacement times the component of force in the direction of displacement, i.e.,

NB

Fs

(b)

Fig. 14–6

UP = 400 N cos 30(2 m) = 692.8 J spring Force Fs. In the initial position the spring is stretched s1 = 0.5 m and in the final position it is stretched s2 = 0.5 m + 2 m = 2.5 m. We require the work to be negative since the force and displacement are opposite to each other. The work of Fs is thus Us = - 3 12(30 N>m)(2.5 m)2 - 12(30 N>m)(0.5 m)2 4 = -90 J

Weight W. Since the weight acts in the opposite sense to its vertical displacement, the work is negative; i.e., UW = -(98.1 N) (2 m sin 30) = -98.1 J Note that it is also possible to consider the component of weight in the direction of displacement; i.e., UW = -(98.1 sin 30 N) (2 m) = -98.1 J Normal Force NB. This force does no work since it is always perpendicular to the displacement. total Work. The work of all the forces when the block is displaced 2 m is therefore UT = 692.8 J - 90 J - 98.1 J = 505 J

30

(a)

Ans.

14

184

Chapter 14

KinetiCs

of a

partiCle: WorK

14.2 ds

s

v

Ft

t

u

2

1

FR  F

Fn

14

n

and

energy

Principle of Work and Energy

Consider the particle in Fig. 14–7, which is located on the path defined relative to an inertial coordinate system. If the particle has a mass m and is subjected to a system of external forces represented by the resultant FR = F, then the equation of motion for the particle in the tangential direction is Ft = mat . Applying the kinematic equation at = v dv>ds and integrating both sides, assuming initially that the particle has a position s = s1 and a speed v = v1 , and later at s = s2 , v = v2 , we have

Fig. 14–7

s2



Ls1

s2



Ls1

y2

Ft ds =

Ly1

mv dv

Ft ds = 12 mv22 - 12 mv21

(14–5)

From Fig. 14–7, note that Ft = Fcos u, and since work is defined from Eq. 14–1, the final result can be written as U1 - 2 = 12 mv22 - 12 mv21

(14–6)

This equation represents the principle of work and energy for the particle. The term on the left is the sum of the work done by all the forces acting on the particle as the particle moves from point 1 to point 2. The two terms on the right side, which are of the form T = 12 mv2, define the particle’s final and initial kinetic energy, respectively. Like work, kinetic energy is a scalar and has units of joules (J). However, unlike work, which can be either positive or negative, the kinetic energy is always positive, regardless of the direction of motion of the particle. When Eq. 14–6 is applied, it is often expressed in the form T1 + U1 - 2 = T2

(14–7)

which states that the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to its final position is equal to the particle’s final kinetic energy. As noted from the derivation, the principle of work and energy represents an integrated form of Ft = mat , obtained by using the kinematic equation at = v dv>ds. As a result, this principle will provide a convenient substitution for Ft = mat when solving those types of kinetic problems which involve force, velocity, and displacement since these quantities are involved in Eq. 14–7. For application, it is suggested that the following procedure be used.

14.2

prinCiple of WorK and energy

185

procedure for analysis Work (Free-Body Diagram). • Establish the inertial coordinate system and draw a free-body diagram of the particle in order to account for all the forces that do work on the particle as it moves along its path. principle of Work and Energy. • Apply the principle of work and energy, T1 + U1 - 2 = T2 .

• The kinetic energy at the initial and final points is always positive, since it involves the speed squared 1 T = 12 mv2 2 .

• A force does work when it moves through a displacement in the direction of the force.

• Work is positive when the force component is in the same sense of direction as its displacement, otherwise it is negative.

• Forces that are functions of displacement must be integrated to

obtain the work. Graphically, the work is equal to the area under the force-displacement curve.

• The work of a weight is the product of the weight magnitude and

the vertical displacement, UW = {Wy. It is positive when the weight moves downwards.

• The work of a spring is of the form Us = 12 ks2, where k is the spring stiffness and s is the stretch or compression of the spring.

Numerical application of this procedure is illustrated in the examples following Sec. 14.3.

If an oncoming car strikes these crash barrels, the car’s kinetic energy will be transformed into work, which causes the barrels, and to some extent the car, to be deformed. By knowing the amount of energy that can be absorbed by each barrel it is possible to design a crash cushion such as this.

14

186

Chapter 14

KinetiCs

of a

partiCle: WorK

14.3

14

and

energy

Principle of Work and Energy for a System of Particles

The principle of work and energy can be extended to include a system of particles isolated within an enclosed region of space as shown in Fig. 14–8. Here the arbitrary ith particle, having a mass mi , is subjected to a resultant external force Fi and a resultant internal force fi which all the other particles exert on the ith particle. If we apply the principle of work and energy to this and each of the other particles in the system, then since work and energy are scalar quantities, the equations can be summed algebraically, which gives T1 + U1 - 2 = T2

(14–8)

In this case, the initial kinetic energy of the system plus the work done by all the external and internal forces acting on the system is equal to the final kinetic energy of the system. If the system represents a translating rigid body, or a series of connected translating bodies, then all the particles in each body will undergo the same displacement. Therefore, the work of all the internal forces will occur in equal but opposite collinear pairs and so it will cancel out. On the other hand, if the body is assumed to be nonrigid, the particles of the body may be displaced along different paths, and some of the energy due to force interactions would be given off and lost as heat or stored in the body if permanent deformations occur. We will discuss these effects briefly at the end of this section and in Sec. 15.4. Throughout this text, however, the principle of work and energy will be applied to problems where direct accountability of such energy losses does not have to be considered.

si n

Fi

i

t fi

Inertial coordinate system

Fig. 14–8

14.3

Work of Friction Caused by Sliding.

A special class of problems will now be investigated which requires a careful application of Eq. 14–8. These problems involve cases where a body slides over the surface of another body in the presence of friction. Consider, for example, a block which is translating a distance s over a rough surface as shown in Fig. 14–9a. If the applied force P just balances the resultant frictional force mkN, Fig. 14–9b, then due to equilibrium a constant velocity v is maintained, and one would expect Eq. 14–8 to be applied as follows: 1 2 2 mv

187

prinCiple of WorK and energy for a system of partiCles v

v P s (a)

W

+ Ps - mkNs = 12 mv2

Indeed this equation is satisfied if P = mkN; however, as one realizes from experience, the sliding motion will generate heat, a form of energy which seems not to be accounted for in the work-energy equation. In order to explain this paradox and thereby more closely represent the nature of friction, we should actually model the block so that the surfaces of contact are deformable (nonrigid).* Recall that the rough portions at the bottom of the block act as “teeth,” and when the block slides these teeth deform slightly and either break off or vibrate as they pull away from “teeth” at the contacting surface, Fig. 14–9c. As a result, frictional forces that act on the block at these points are displaced slightly, due to the localized deformations, and later they are replaced by other frictional forces as other points of contact are made. At any instant, the resultant F of all these frictional forces remains essentially constant, i.e., mkN; however, due to the many localized deformations, the actual displacement s of mkN is not the same as the displacement s of the applied force P. Instead, s will be less than s (s 6 s), and therefore the external work done by the resultant frictional force will be mkNs and not mkNs. The remaining amount of work, mkN(s - s), manifests itself as an increase in internal energy, which in fact causes the block’s temperature to rise. In summary then, Eq. 14–8 can be applied to problems involving sliding friction; however, it should be fully realized that the work of the resultant frictional force is not represented by mkNs; instead, this term represents both the external work of friction (mkNs) and internal work [mkN(s - s)] which is converted into various forms of internal energy, such as heat.†

*See Chapter 8 of Engineering Mechanics: Statics. †See B. A. Sherwood and W. H. Bernard, “Work and Heat Transfer in the Presence of Sliding Friction,” Am. J. Phys. 52, 1001 (1984).

P

14 P

F  mkN N (b)

(c)

Fig. 14–9

188

Chapter 14

KinetiCs

of a

partiCle: WorK

and

energy

14.2

ExamplE

6 m/s A

s 10

The 1750-kg automobile shown in Fig. 14–10a travels down the 10° inclined road at a speed of 6 m>s. If the driver jams on the brakes, causing his wheels to lock, determine how far s the tires skid on the road. The coefficient of kinetic friction between the wheels and the road is mk = 0.5.

(a)

14 s

1750(9.81) N

10 FA 10 NA (b)

Fig. 14–10

Solution This problem can be solved using the principle of work and energy, since it involves force, velocity, and displacement. Work (Free-body Diagram). As shown in Fig. 14–10b, the normal force NA does no work since it never undergoes displacement along its line of action. The weight, 1750(9.81)N, is displaced s sin 10° and does positive work. Why? The frictional force FA does both external and internal work when it undergoes a displacement s. This work is negative since it is in the opposite sense of direction to the displacement. Applying the equation of equilibrium normal to the road, we have + aFn = 0;

NA - 1750(9.81) cos 10 N = 0

NA = 16 906.7 N

Thus, FA = mk NA = 0.5 (16 906.7 N) = 8453.35 N principle of Work and energy. T1 + U192 = T2 1 (1750 kg)(6 m>s)2 + 1750(9.81)N (s sin 10) - (8453.35 N)s = 0 2 Solving for s yields s = 5.76 m

Ans.

NoTE: If this problem is solved by using the equation of motion, two steps are involved. First, from the free-body diagram, Fig. 14–10b, the equation of motion is applied along the incline. This yields

+ bFs = mas;

1750(9.81) sin 10 N - 8453.35 N = (1750 kg) a a = -3.127 m>s2

Then, since a is constant, we have

1 +b2

v2 = v20 + 2ac(s - s0); (0)2 = (6 m>s)2 + 2( -3.127 m>s2)(s - 0) s = 5.76 m

Ans.

prinCiple of WorK and energy for a system of partiCles

14.3

189

14.3

ExamplE

For a short time the crane in Fig. 14–11a lifts the 2.50-Mg beam with a force of F = (28 + 3s2) kN. Determine the speed of the beam when it has risen s = 3 m. Also, how much time does it take to attain this height starting from rest? Solution We can solve part of this problem using the principle of work and energy since it involves force, velocity, and displacement. Kinematics must be used to determine the time. Note that at s = 0, F = 28(103)N 7 W = 2.50(103)(9.81)N, so motion will occur.

14

Work (Free-body Diagram). As shown on the free-body diagram, Fig. 14–11b, the lifting force F does positive work, which must be determined by integration since this force is a variable. Also, the weight is constant and will do negative work since the displacement is upward. (a)

principles of Work and energy. T1 + U192 = T2 0 +

L0

F

s 2

3

3

(28 + 3s )(10 ) ds - (2.50)(10 )(9.81)s =

1 3 2 2 (2.50)(10 )v

28(103)s + (103)s3 - 24.525(103)s = 1.25(103)v2 1

v = (2.78s + 0.8s3)2

(1)

When s = 3 m, Ans.

v = 5.47 m>s

Kinematics. Since we were able to express the velocity as a function of displacement, the time can be determined using v = ds>dt. In this case, 1

(2.78s + 0.8s3)2 = 3

t =

Fig. 14–11

ds dt

L0 (2.78s + 0.8s3)2 ds

1

The integration can be performed numerically using a pocket calculator. The result is t = 1.79 s

2.50 (103)(9.81) N (b)

Ans.

NoTE: The acceleration of the beam can be determined by integrating

Eq. (1) using v dv = a ds, or more directly, by applying the equation of motion, F = ma.

190

Chapter 14

ExamplE

KinetiCs

of a

partiCle: WorK

and

energy

14.4 The platform P, shown in Fig. 14–12a, has negligible mass and is tied down so that the 0.4-m-long cords keep a 1-m-long spring compressed 0.6 m when nothing is on the platform. If a 2-kg block is placed on the platform and released from rest after the platform is pushed down 0.1 m, Fig. 14–12b, determine the maximum height h the block rises in the air, measured from the ground.

14

P

h

k  200 N/m 0.4 m

0.3 m

(a)

(b)

Fig. 14–12

19.62 N

Solution Work (Free-body Diagram). Since the block is released from rest and later reaches its maximum height, the initial and final velocities are zero. The free-body diagram of the block when it is still in contact with the platform is shown in Fig. 14–12c. Note that the weight does negative work and the spring force does positive work. Why? In particular, the initial compression in the spring is s1 = 0.6 m + 0.1 m = 0.7 m. Due to the cords, the spring’s final compression is s2 = 0.6 m (after the block leaves the platform). The bottom of the block rises from a height of (0.4 m - 0.1 m) = 0.3 m to a final height h. principle of Work and energy. T1 + U192 = T2 1 2 2 mv1

Fs

(c)

+

5 -1

1 2 2 ks2

-

1 2 2 ks1

2

- W y 6 = 12 mv22

Note that here s1 = 0.7 m 7 s2 = 0.6 m and so the work of the spring as determined from Eq. 14–4 will indeed be positive once the calculation is made. Thus, 0 +

5 - 3 12(200 N>m)(0.6 m)2

Solving yields

- 12(200 N>m)(0.7 m)2 4

- (19.62 N)[h - (0.3 m)] 6 = 0

h = 0.963 m

Ans.

PrinciPle of Work and energy for a SyStem of ParticleS

14.3

ExamplE

191

14.5

The 40-kg boy in Fig. 14–13a slides down the smooth water slide. If he starts from rest at A, determine his speed when he reaches B and the normal reaction the slide exerts on the boy at this position. y y  0.075x2

A

14

7.5 m B

Please refer to the Companion Website for the animation: A Skier Showing the Work Done at Different Stages (from Positions 1, 2 to 3)

x 10 m

(a)

Solution Work (Free-Body Diagram). As shown on the free-body diagram, Fig. 14–13b, there are two forces acting on the boy as he goes down the slide. Note that the normal force does no work.

n

40(9.81) N u u

Principle of Work and Energy.

t

Nb

TA + UA - B = TB (b)

0 + (40(9.81)N) (7.5 m) = 12(40 kg)v2B Ans.

vB = 12.13 m>s = 12.1 m>s

Equation of Motion. Referring to the free-body diagram of the boy when he is at B, Fig. 14–13c, the normal reaction NB can now be obtained by applying the equation of motion along the n axis. Here the radius of curvature of the path is

rB =

dy 2 3>2 b d dx =  d2y>dx2 

c1 + a

40(9.81) N t

31

+ (0.15x)2 4 3>2  0.15 

3

= 6.667 m NB

x=0

Thus,

(c) 2

+ c Fn = man;

n

NB - 40(9.81) N = 40 kg¢ NB = 1275.3 N = 1.28 kN

(12.13 m>s) ≤ 6.667 m

Fig. 14–13

Ans.

192

Chapter 14

KinetiCs

partiCle: WorK

and

energy

14.6

ExamplE

Datum sB

sA

B 100 kg

14

A 10 kg (a)

T

of a

R1

R2

Blocks A and B shown in Fig. 14–14a have a mass of 10 kg and 100 kg, respectively. Determine the distance B travels when it is released from rest to the point where its speed becomes 2 m>s. Solution This problem may be solved by considering the blocks separately and applying the principle of work and energy to each block. However, the work of the (unknown) cable tension can be eliminated from the analysis by considering blocks A and B together as a single system. Work (Free-body Diagram). As shown on the free-body diagram of the system, Fig. 14–14b, the cable force T and reactions R1 and R2 do no work, since these forces represent the reactions at the supports and consequently they do not move while the blocks are displaced. The weights both do positive work if we assume both move downward, in the positive sense of direction of sA and sB. principle of Work and energy. Realizing the blocks are released from rest, we have

B

T1 + U192 = T2 981 N A

98.1 N (b)

Fig. 14–14

5

1 2 2 mA(vA)1

+

1 2 2 mB(vB)1

6

+ 5 WA sA + WB sB 6 =

5 0 + 0 6 + 5 98.1 N (sA) + 981 N (sB) 6 =

5 12(10 kg)(vA)22

5 12 mA(vA)22

+ 12(100 kg)(2 m>s)2 6

+ 12 mB(vB)22 6

(1)

Kinematics. Using methods of kinematics, as discussed in Sec. 12.9, it may be seen from Fig. 14–14a that the total length l of all the vertical segments of cable may be expressed in terms of the position coordinates sA and sB as sA + 4sB = l Hence, a change in position yields the displacement equation sA + 4 sB = 0 sA = -4 sB Here we see that a downward displacement of one block produces an upward displacement of the other block. Note that sA and sB must have the same sign convention in both Eqs. 1 and 2. Taking the time derivative yields vA = -4vB = -4(2 m>s) = -8 m>s (2) Retaining the negative sign in Eq. 2 and substituting into Eq. 1 yields sB = 0.883 m T

Ans.

prinCiple of WorK and energy for a system of partiCles

14.3

193

preliMiNary prObleMs P14–1. Determine the work of the force when it displaces 2 m.

k  10 N/m 2m

500 N Spring is originally compressed 3 m.

5

4 3

2m

14

(f)

100 N 5

(a)

3

4

98.1 N 2m

2m

(g) Prob. P14–1 (b) P14–2. Determine the kinetic energy of the 10-kg block.

2m 2

F  (6 s ) N

(c)

2 m/s

100 N

2m 30 3

(a)

5 4

(d) 6 m/s

F (N)

F 3

60

5 4

2m

3m

20

1

(e)

2

s (m)

(b) Prob. P14–2

194

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and

energy

FuNDaMeNtal prObleMs F14–1. The spring is placed between the wall and the 10-kg block. If the block is subjected to a force of F = 500 N, determine its velocity when s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth. 14

500 N

Prob. F14–1

F14–4. The 1.8-Mg dragster is traveling at 125 m>s when the engine is shut off and the parachute is released. If the drag force of the parachute can be approximated by the graph, determine the speed of the dragster when it has traveled 400 m. FD (kN)

5

3

4

s k  500 N/m

50 20

F14–2. If the motor exerts a constant force of 300 N on the cable, determine the speed of the 20-kg crate when it travels s = 10 m up the plane, starting from rest. The coefficient of kinetic friction between the crate and the plane is mk = 0.3. M

s (m) 400

F14–5. When s = 0.6 m, the spring is unstretched and the 10-kg block has a speed of 5 m>s down the smooth plane. Determine the distance s when the block stops.

s  10 m

s

k  200 N/m 5 m/s

30

A

Prob. F14–4

F  100 N

Prob. F14–2 F14–3. If the motor exerts a force of F = (600 + 2s2) N on the cable, determine the speed of the 100-kg crate when it rises to s = 15 m. The crate is initially at rest on the ground. M

C

30

Prob. F14–5 F14–6. The 2.5-kg collar is pulled by a cord that passes around a small peg at C. If the cord is subjected to a constant force of F = 50 N, and the collar is at rest when it is at A, determine its speed when it reaches B. Neglect friction. 2m

A

B 1.5 m

15 m

s

Prob. F14–3

C F  50 N

Prob. F14–6

14.3

195

prinCiple of WorK and energy for a system of partiCles

prObleMs 14–1. The 20-kg crate is subjected to a force having a constant direction and a magnitude F = 100 N. When s = 15 m, the crate is moving to the right with a speed of 8 m>s. Determine its speed when s = 25 m. The coefficient of kinetic friction between the crate and the ground is mk = 0.25. F

*14–4. Determine the required height h of the roller coaster so that when it is essentially at rest at the crest of the hill A, it will reach a speed of 100 km>h when it comes to the bottom B. Also, what should be the minimum radius of curvature r for the track at B so that the passengers do not experience a normal force greater than 4mg = (39.24m) N? Neglect the size of the car and passenger. 14

30 A

Prob. 14–1 h

14–2. The crate, which has a mass of 100 kg, is subjected to the action of the two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m>s. The coefficient of kinetic friction between the crate and the surface is mk = 0.2.

r B

Prob. 14–4

1000 N

800 N 5

30

3 4

Prob. 14–2 14–3. The 100-kg crate is subjected to the forces shown. If it is originally at rest, determine the distance it slides in order to attain a speed of v = 8 m>s. The coefficient of kinetic friction between the crate and the surface is mk = 0.2.

14–5. For protection, the barrel barrier is placed in front of the bridge pier. If the relation between the force and deflection of the barrier is F = [800(103)x1> 2] N, where x is in m, determine the car’s maximum penetration in the barrier. The car has a mass of 2 Mg and it is traveling with a speed of 20 m>s just before it hits the barrier.

F (N)

F  800(103) x1/2

500 N

400 N 30

x (m)

45

Prob. 14–3

Prob. 14–5

196

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of a

partiCle: WorK

14–6. The force of F = 50 N is applied to the cord when s = 2 m. If the 6-kg collar is orginally at rest, determine its velocity at s = 0. Neglect friction.

F

14 1.5 m

and

energy

*14–8. As indicated by the derivation, the principle of work and energy is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which rests on the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block if it has an initial speed of 5 m>s and travels 10 m, both directed to the right and measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x  axis and moving at a constant velocity of 2 m>s relative to A. Hint: The distance the block travels will first have to be computed for observer B before applying the principle of work and energy.

A

s A

Prob. 14–6

B

x x¿

2 m/s 5 m/s

14–7. A force of F = 250 N is applied to the end at B. Determine the speed of the 10-kg block when it has moved 1.5 m, starting from rest.

6N

10 m

Prob. 14–8

14–9. When the driver applies the brakes of a light truck traveling 40 km>h, it skids 3 m before stopping. How far will the truck skid if it is traveling at 80 km>h when the brakes are applied?

B

A

F

Prob. 14–7

Prob. 14–9

14.3

14–10. The “air spring” A is used to protect the support B and prevent damage to the conveyor-belt tensioning weight C in the event of a belt failure D. The force developed by the air spring as a function of its deflection is shown by the graph. If the block has a mass of 20 kg and is suspended a height d = 0.4 m above the top of the spring, determine the maximum deformation of the spring in the event the conveyor belt fails. Neglect the mass of the pulley and belt.

*14–12. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed yB when he reaches B. Also, find the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass of 70 kg.

14

A

F (N)

D

197

prinCiple of WorK and energy for a system of partiCles

50 m

1500

4m

B

s

C s (m)

0.2

d

C

30

A

Prob. 14–12

B

Prob. 14–10 14–11. The force F, acting in a constant direction on the 20-kg block, has a magnitude which varies with the position s of the block. Determine how far the block must slide before its velocity becomes 15 m>s. When s = 0 the block is moving to the right at v = 6 m>s. The coefficient of kinetic friction between the block and surface is mk = 0.3.

v F (N)

F

14–13. Design considerations for the bumper B on the 5-Mg train car require use of a nonlinear spring having the load-deflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to 0.2 m when the car, traveling at 4 m>s, strikes the rigid stop. Neglect the mass of the car wheels.

F (N)

F  ks2

s (m) F  50s 1/2

B s (m)

Prob. 14–11

Prob. 14–13

198

Chapter 14

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of a

partiCle: WorK

14–14. The 8-kg cylinder A and 3-kg cylinder B are released from rest. Determine the speed of A after it has moved 2 m starting from rest. Neglect the mass of the cord and pulleys.

and

energy

14–17. A small box of mass m is given a speed of v = 241 gr at the top of the smooth half cylinder. Determine the angle u at which the box leaves the cylinder.

14–15. Cylinder A has a mass of 3 kg and cylinder B has a mass of 8 kg. Determine the speed of A after it has moved 2  m starting from rest. Neglect the mass of the cord and pulleys.

14 A u

r

O

Prob. 14–17 A B

Probs. 14–14/15

14–18. If the cord is subjected to a constant force of F = 300 N and the 15-kg smooth collar starts from rest at A, determine the velocity of the collar when it reaches point B. Neglect the size of the pulley.

*14–16. The collar has a mass of 20 kg and is supported on the smooth rod. The attached springs are undeformed when d = 0.5 m. Determine the speed of the collar after the applied force F = 100 N causes it to be displaced so that d = 0.3 m. When d = 0.5 m the collar is at rest.

200 mm

k' = 15 N/m

B

F = 100 N

C 30 200 mm F  300 N

60°

d

200 mm

300 mm

k = 25 N/m A

Prob. 14–16

Prob. 14–18

14.3

prinCiple of WorK and energy for a system of partiCles

14–19. If the force exerted by the motor M on the cable is 250 N, determine the speed of the 100-kg crate when it is hoisted to s = 3 m.The crate is at rest when s = 0.

199

14–21. The steel ingot has a mass of 1800 kg. It travels along the conveyor at a speed v = 0.5 m>s when it collides with the “nested” spring assembly. If the stiffness of the outer spring is kA = 5 kN>m determine the required stiffness kB of the inner spring so that the motion of the ingot is stopped at the moment the front, C, of the ingot is 0.3 m from the wall.

M

14

0.5 m 0.45 m kA

kB

A

s

C

B

Prob. 14–21

Prob. 14–19

*14–20. When a 7-kg projectile is fired from a cannon barrel that has a length of 2 m, the explosive force exerted on the projectile, while it is in the barrel, varies in the manner shown. Determine the approximate muzzle velocity of the projectile at the instant it leaves the barrel. Neglect the effects of friction inside the barrel and assume the barrel is horizontal.

14–22. The 1.5-kg block slides along a smooth plane and strikes a nonlinear spring with a speed of v = 4 m>s. The spring is termed “nonlinear” because it has a resistance of Fs = ks2, where k = 900 N>m2. Determine the speed of the block after it has compressed the spring s = 0.2 m.

F (MN) 15 v 10

k

5

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

Prob. 14–20

s (m)

Prob. 14–22

200

Chapter 14

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of a

partiCle: WorK

14–23. A car is equipped with a bumper B designed to absorb collisions. The bumper is mounted to the car using pieces of flexible tubing T. Upon collision with a rigid barrier at A, a constant horizontal force F is developed which causes a car deceleration of 3g = 29.43 m>s2 (the highest safe deceleration for a passenger without a seatbelt). If the car and passenger have a total mass of 1.5 Mg and the car is initially coasting with a speed of 1.5 m>s, determine the magnitude of F needed to stop the car and the deformation x of the bumper tubing.

and

energy

14–25. The 12-kg block has an initial speed of v0 = 4 m when it is midway between springs A and B. After striking spring B, it rebounds and slides across the horizontal plane toward spring A, etc. If the coefficient of kinetic friction between the plane and the block is mk = 0.4, determine the total distance traveled by the block before it comes to rest. 0.6 m

kA  100 N/m

kB  300 N/m

0.3 m v0  4 m/s

14 A

B

Prob. 14–25 1.5 m/s A

14–26. The 8-kg block is moving with an initial speed of 5 m>s. If the coefficient of kinetic friction between the block and plane is mk = 0.25, determine the compression in the spring when the block momentarily stops.

B T 5 m/s

T T

2m

B

kA  200 N/m

A

Prob. 14–23 Prob. 14–26 *14–24. The catapulting mechanism is used to propel the 10-kg slider A to the right along the smooth track. The propelling action is obtained by drawing the pulley attached to rod BC rapidly to the left by means of a piston P. If the piston applies a constant force F = 20 kN to rod BC such that it moves it 0.2 m, determine the speed attained by the slider if it was originally at rest. Neglect the mass of the pulleys, cable, piston, and rod BC.

14–27. Marbles having a mass of 5 g are dropped from rest at A through the smooth glass tube and accumulate in the can at C. Determine the placement R of the can from the end of the tube and the speed at which the marbles fall into the can. Neglect the size of the can. A

B 3m

A

2m P F

B

C

C R

Prob. 14–24

Prob. 14–27

14.3

201

prinCiple of WorK and energy for a system of partiCles

*14–28. The collar has a mass of 20 kg and slides along the smooth rod. Two springs are attached to it and the ends of the rod as shown. If each spring has an uncompressed length of 1 m and the collar has a speed of 2 m>s when s = 0, determine the maximum compression of each spring due to the back-and-forth (oscillating) motion of the collar.

14–30. The 0.5-kg ball is fired up the smooth vertical circular track using the spring plunger. The plunger keeps the spring compressed 0.08 m when s = 0. Determine how far s it must be pulled back and released so that the ball will begin to leave the track when u = 135.

B

14 135 u s

s

1.5 m

A k

kA  50 N/m

kB  100 N/m

1m

1m 0.25 m

Prob. 14–28

14–29. The train car has a mass of 10 Mg and is traveling at 5 m>s when it reaches A. If the rolling resistance is 1>100 of the weight of the car, determine the compression of each spring when the car is momentarily brought to rest.

500 N/m

Prob. 14–30 14–31. The “flying car” is a ride at an amusement park which consists of a car having wheels that roll along a track mounted inside a rotating drum. By design the car cannot fall off the track, however motion of the car is developed by applying the car’s brake, thereby gripping the car to the track and allowing it to move with a constant speed of the track, vt = 3 m>s. If the rider applies the brake when going from B to A and then releases it at the top of the drum, A, so that the car coasts freely down along the track to B (u = p rad), determine the speed of the car at B and the normal reaction which the drum exerts on the car at B. Neglect friction during the motion from A to B. The rider and car have a total mass of 250 kg and the center of mass of the car and rider moves along a circular path having a radius of 8 m. A vt

u

8m 300 kN/m 500 kN/m

vA

30 m

5 m/s

A

B

Prob. 14–29

Prob. 14–31

202

Chapter 14

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partiCle: WorK

*14–32. The man at the window A wishes to throw the 30-kg sack on the ground. To do this he allows it to swing from rest at B to point C, when he releases the cord at u = 30. Determine the speed at which it strikes the ground and the distance R.

and

energy

14–34. The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s velocity is yA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs.

vA

14

2.5 m/s A

A B

3m

8m 8m

B

u 16 m

30

C

Prob. 14–34

D R

Prob. 14–32 14–33. The cyclist travels to point A, pedaling until he reaches a speed yA = 4 m>s. He then coasts freely up the curved surface. Determine how high he reaches up the  surface before he comes to a stop. Also, what are the resultant normal force on the surface at this point and his acceleration? The total mass of the bike and man is 75 kg. Neglect friction, the mass of the wheels, and the size of the bicycle.

14–35. The block has a mass of 0.8 kg and moves within the smooth vertical slot. If it starts from rest when the attached spring is in the unstretched position at A, determine the constant vertical force F which must be applied to the cord so that the block attains a speed vB = 2.5 m>s when it reaches B; sB = 0.15 m. Neglect the size and mass of the pulley. Hint: The work of F can be determined by finding the difference l in cord lengths AC and BC and using UF = F l. 0.3 m C

y

0.4 m C

B 4m

x1/2

y1/2

B y

x

45

sB

2

A A

F k  100 N/m

x

4m

Prob. 14–33

Prob. 14–35

203

prinCiple of WorK and energy for a system of partiCles

14.3

*14–36. If the 60-kg skier passes point A with a speed of 5 m>s, determine his speed when he reaches point B. Also find the normal force exerted on him by the slope at this point. Neglect friction. y

14–39. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed vB when he reaches B. Also, find the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier's size. He has a mass 75 kg.

y  (0.025x2  5) m A

14

A 15 m

B

50 m 4m

B x

s

Prob. 14–36 14–37. The spring in the toy gun has an unstretched length of 100 mm. It is compressed and locked in the position shown. When the trigger is pulled, the spring unstretches 12.5 mm, and the 20-g ball moves along the barrel. Determine the speed of the ball when it leaves the gun. Neglect friction.

C

30

Prob. 14–39

50 mm k

2 kN/m

D

*14–40. If the 75-kg crate starts from rest at A, determine its speed when it reaches point B. The cable is subjected to a constant force of F = 300 N. Neglect friction and the size of the pulley.

150 mm A B

14–41. If the 75-kg crate starts from rest at A, and its speed is 6 m>s when it passes point B, determine the constant force F exerted on the cable. Neglect friction and the size of the pulley. Prob. 14–37 14–38. If the track is to be designed so that the passengers of the roller coaster do not experience a normal force equal to zero or more than 4 times their weight, determine the limiting heights hA and hC so that this does not occur. The roller coaster starts from rest at position A. Neglect friction.

C 30 F

A C hC

6m

rC  20 m rB  15 m

hA

B A

B 6m

Prob. 14–38

2m

Probs. 14–40/41

204

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of a

partiCle: WorK

14.4

and

energy

Power and Efficiency

Power. The term “power” provides a useful basis for choosing the type of motor or machine which is required to do a certain amount of work in a given time. For example, two pumps may each be able to empty a reservoir if given enough time; however, the pump having the larger power will complete the job sooner. The power generated by a machine or engine that performs an amount of work dU within the time interval dt is therefore

14

P =

dU dt

(14–9)

If the work dU is expressed as dU = F # dr, then

P =

dU F # dr dr = = F# dt dt dt

or P = F#v

(14–10)

Hence, power is a scalar, where in this formulation v represents the velocity of the particle which is acted upon by the force F. The basic unit of power used in the SI system is the watt (W). These units are defined as The power output of this locomotive comes from the driving frictional force developed at its wheels. It is this force that overcomes the frictional resistance of the cars in tow and is able to lift the weight of the train up the grade.

1 W = 1 J>s = 1 N # m>s

Efficiency. The mechanical efficiency of a machine is defined as the

ratio of the output of useful power produced by the machine to the input of power supplied to the machine. Hence,

e =

power output power input

(14–11)

14.4

poWer and effiCienCy

205

If energy supplied to the machine occurs during the same time interval at which it is drawn, then the efficiency may also be expressed in terms of the ratio e =

energy output energy input

(14–12)

Since machines consist of a series of moving parts, frictional forces will always be developed within the machine, and as a result, extra energy or power is needed to overcome these forces. Consequently, power output will be less than power input and so the efficiency of a machine is always less than 1. The power supplied to a body can be determined using the following procedure.

procedure for analysis • First determine the external force F acting on the body which causes the motion. This force is usually developed by a machine or engine placed either within or external to the body.

• If the body is accelerating, it may be necessary to draw its free-

body diagram and apply the equation of motion (F = ma) to determine F.

• Once F and the velocity v of the particle where F is applied have

been found, the power is determined by multiplying the force magnitude with the component of velocity acting in the direction of F, (i.e., P = F # v = Fv cos u).

• In some problems the power may be found by calculating the work done by F per unit of time (Pavg = U> t,).

F v

The power requirement of this hoist depends upon the vertical force F that acts on the elevator and causes it to move upward. If the velocity of the elevator is v, then the power output is P = F # v.

14

206

Chapter 14

ExamplE

KinetiCs

of a

partiCle: WorK

and

energy

14.7 The man in Fig. 14–15a pushes on the 50-kg crate with a force of F = 150 N. Determine the power supplied by the man when t = 4 s. The coefficient of kinetic friction between the floor and the crate is mk = 0.2. Initially the create is at rest.

14

y F  150 N 3

50 (9.81) N

5 4

F  150 N 3

a x

5 4

Fƒ  0.2 N N (b)

(a)

Fig. 14–15

Solution To determine the power developed by the man, the velocity of the 150-N force must be obtained first. The free-body diagram of the crate is shown in Fig. 14–15b. Applying the equation of motion, + c Fy = may;

N -

1 35 2 150 N

- 50(9.81) N = 0

N = 580.5 N

+ Fx = max; S

1 45 2 150 N

- 0.2(580.5 N) = (50 kg)a a = 0.078 m>s2

The velocity of the crate when t = 4 s is therefore + ) (S

v = v0 + act v = 0 + (0.078 m>s2)(4 s) = 0.312 m>s

The power supplied to the crate by the man when t = 4 s is therefore P = F # v = Fxv = = 37.4 W

1 45 2 (150 N)(0.312 m>s)

Ans.

14.4

ExamplE

207

Power and efficiency

14.8

The motor M of the hoist shown in Fig. 14–16a lifts the 35-kg crate C so that the acceleration of point P is 1.2 m>s2. Determine the power that must be supplied to the motor at the instant P has velocity of 0.6 m>s. Neglect the mass of the pulley and cable and take e = 0.85. Solution In order to find the power output of the motor, it is first necessary to determine the tension in the cable since this force is developed by the motor. From the free-body diagram, Fig. 14–16b, we have

Datum sP Datum

P

M

sC

14

C

+T

Fy = may;

-2T + 35(9.81) N = (35 kg) ac

(1)

The acceleration of the crate can be obtained by using kinematics to relate it to the known acceleration of point P, Fig. 14–16a. Using the methods of absolute dependent motion, the coordinates sC and sP can be related to a constant portion of cable length l which is changing in the vertical and horizontal directions. We have 2sC + sP = l. Taking the second time derivative of this equation yields 2aC = -aP 2

(2) 2

2

Since aP = +1.2 m>s , then aC = -(1.2 m>s )>2 = -0.6 m>s . What does the negative sign indicate? Substituting this result into Eq. 1 and retaining the negative sign since the acceleration in both Eq. 1 and Eq. 2 was considered positive downward, we have

aC y

35(9.81) N

Fig. 14–16

T = 182.2 N The power output required to draw the cable in at a rate of 0.6 m>s is therefore P = T # v = (182.2 N)(0.6 m/s) = 109.3 W This power output requires that the motor provide a power input of 1 (power output) e 1 = (109.3 W) = 129 W 0.85

2T

(b)

-2T + 35(9.81) N = (35 kg)( -0.6 m/s2)

power input =

Ans.

Note: Since the velocity of the crate is constantly changing, the

power requirement is instantaneous.

(a)

208

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and

energy

FuNDaMeNtal prObleMs F14–7. If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest.

F14–10. The coefficient of kinetic friction between the 20-kg block and the inclined plane is mk = 0.2. If the block is traveling up the inclined plane with a constant velocity v = 5 m>s, determine the power of force F. F

F  30 N 5 4

14

3

Prob. F14–7 30

F14–8. If F = (10s) N, where s is in meters, and the contact surface between the block and the ground is smooth, determine the power of force F when s = 5 m. When s = 0, the 20-kg block is moving at v = 1 m>s.

Prob. F14–10 F14–11. If the 50-kg load A is hoisted by motor M so that the load has a constant velocity of 1.5 m>s, determine the power input to the motor, which operates at an efficiency e = 0.8. M

F  (10 s) N

s

Prob. F14–8 F14–9. If the motor winds in the cable with a constant speed of v = 3 m>s, determine the power supplied to the motor. The load weighs 100 N and the efficiency of the motor is e = 0.8. Neglect the mass of the pulleys.

Prob. F14–11

B A

1.5 m/s

F14–12. At the instant shown, point P on the cable has a velocity vP = 12 m>s, which is increasing at a rate of aP = 6 m>s2. Determine the power input to motor M at this instant if it operates with an efficiency e = 0.8. The mass of block A is 50 kg.

D

C

A

P

12 m/s 6 m/s2

v  3 m/s M

E M

Prob. F14–9

A

Prob. F14–12

14.4

poWer and effiCienCy

209

prObleMs 14–42. A spring having a stiffness of 5 kN>m is compressed 400 mm. The stored energy in the spring is used to drive a machine which requires 90 W of power. Determine how long the spring can supply energy at the required rate. 14–43. To dramatize the loss of energy in an automobile, consider a car having a weight of 25 000 N that is traveling at 56 km>h. If the car is brought to a stop, determine how long a 100-W light bulb must burn to expend the same amount of energy. *14–44. If the engine of a 1.5-Mg car generates a constant power of 15 kW, determine the speed of the car after it has traveled a distance of 200 m on a level road starting from rest. Neglect friction. 14–45. If the engine of a 1.5-Mg car generates a constant power of 15 kW, determine the speed of the car after it has traveled a distance of 200 m on a level road starting from rest. Neglect friction.

14–46. The 2-Mg car increases its speed uniformly from rest to 25 m>s in 30 s up the inclined road. Determine the maximum power that must be supplied by the engine, which operates with an efficiency of e = 0.8. Also, find the average power supplied by the engine.

*14–48. An automobile having a mass of 2 Mg travels up a 7° slope at a constant speed of v = 100 km>h. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the automobile has an efficiency e = 0.65. 14

7

Prob. 14–48

14–49. A rocket having a total mass of 8 Mg is fired vertically from rest. If the engines provide a constant thrust of T = 300 kN, determine the power output of the engines as a function of time. Neglect the effect of drag resistance and the loss of fuel mass and weight.

1 10

Prob. 14–46

14–47. A car has a mass m and accelerates along a horizontal straight road from rest such that the power is always a constant amount P. Determine how far it must travel to reach a speed of y.

T

300 kN

Prob. 14–49

210

Chapter 14

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14–50. The sports car has a mass of 2.3 Mg, and while it is traveling at 28 m>s the driver causes it to accelerate at 5 m>s2. If the drag resistance on the car due to the wind is FD = (0.3y2) N, where y is the velocity in m>s, determine the power supplied to the engine at this instant. The engine has a running efficiency of e = 0.68.

and

energy

14–53. The 50-kg crate is hoisted up the 30° incline by the pulley system and motor M. If the crate starts from rest and, by constant acceleration, attains a speed of 4 m>s after traveling 8 m along the plane, determine the power that must be supplied to the motor at the instant. Neglect friction along the plane. The motor has an efficiency of e = 0.74.

14 14–51. The sports car has a mass of 2.3 Mg and accelerates at 6 m>s2, starting from rest. If the drag resistance on the car due to the wind is FD = (10v) N, where v is the velocity in m>s, determine the power supplied to the engine when t = 5 s. The engine has a running efficiency of e = 0.68.

M

FD

30

Prob. 14–53 Probs. 14–50/51

*14–52. A motor hoists a 60-kg crate at a constant velocity to a height of h = 5 in 2 s. If the indicated power of the motor is 3.2 kW, determine the motor’s efficiency.

14–54. The 500-kg elevator starts from rest and travels upward with a constant acceleration ac = 2 m>s2. Determine the power output of the motor M when t = 3 s. Neglect the mass of the pulleys and cable.

M

E

h

Prob. 14–52

Prob. 14–54

14.4 14–55. The escalator steps move with a constant speed of 0.6 m>s. If the steps are 125 mm high and 250 mm in length, determine the power of a motor needed to lift an average mass of 150 kg per step. There are 32 steps. *14–56. If the escalator in Prob. 14–55 is not moving, determine the constant speed at which a man having a mass of 80 kg must walk up the steps to generate 100 W of power—the same amount that is needed to power a standard light bulb.

poWer and effiCienCy

211

14–58. The crate has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are ms = 0.3 and mk = 0.2, respectively. If the motor M supplies a cable force of F = (8t2 + 20) N, where t is in seconds, determine the power output developed by the motor when t = 5 s.

M

14 250 mm 125 mm v  0.6 m/s

4m

Prob. 14–58

14–59. The material hoist and the load have a total mass of 800 kg and the counterweight C has a mass of 150 kg. At a given instant, the hoist has an upward velocity of 2 m>s and an acceleration of 1.5 m>s2. Determine the power generated by the motor M at this instant if it operates with an efficiency of e = 0.8.

Probs. 14–55/56 14–57. The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of e = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of vE = 4 m>s.

*14–60. The material hoist and the load have a total mass of 800 kg and the counterweight C has a mass of 150 kg. If the upward speed of the hoist increases uniformly from 0.5  m>s to 1.5 m>s in 1.5 s, determine the average power generated by the motor M during this time. The motor operates with an efficiency of e = 0.8.

M M

C vE E

C

Prob. 14–57

Probs. 14–59/60

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14–61. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same direction as the force varies with time as shown in the second graph. Determine the power applied as a function of time and the work done in t = 0.3 s.

and

energy

*14–64. The rocket sled has a mass of 4 Mg and travels from rest along the horizontal track for which the coefficient of kinetic friction is mk = 0.20. If the engine provides a constant thrust T = 150 kN, determine the power output of the engine as a function of time. Neglect the loss of fuel mass and air resistance.

14–62. An athlete pushes against an exercise machine with a force that varies with time as shown in the first graph. Also, the velocity of the athlete’s arm acting in the same 14 direction as the force varies with time as shown in the second graph. Determine the maximum power developed during the 0.3-second time period.

v T F (N)

800

Prob. 14–64

0.2

0.3

t (s)

v (m/s)

20

0.3

t (s)

Probs. 14–61/62

14–65. The block has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. If a force F = (60t2) N, where t is in seconds, is applied to the cable, determine the power developed by the force when t = 5 s. Hint: First determine the time needed for the force to cause motion.

14–63. If the jet on the dragster supplies a constant thrust of T = 20 kN, determine the power generated by the jet as a function of time. Neglect drag and rolling resistance, and the loss of fuel. The dragster has a mass of 1 Mg and starts from rest.

T

F

Prob. 14–63

Prob. 14–65

14.5

14.5

213

Conservative forCes and potential energy

Conservative Forces and Potential Energy

Conservative Force.

If the work of a force is independent of the path and depends only on the force’s initial and final positions on the path, then we can classify this force as a conservative force. Examples of conservative forces are the weight of a particle and the force developed by a spring. The work done by the weight depends only on the vertical displacement of the weight, and the work done by a spring force depends only on the spring’s elongation or compression. In contrast to a conservative force, consider the force of friction exerted on a sliding object by a fixed surface. The work done by the frictional force depends on the path—the longer the path, the greater the work. Consequently, frictional forces are nonconservative. The work is dissipated from the body in the form of heat.

Vg  Wy

W

y Datum

Vg  0 y

W

Vg  Wy

Energy.

Energy is defined as the capacity for doing work. For example, if a particle is originally at rest, then the principle of work and energy states that U1 S 2 = T2. In other words, the kinetic energy is equal to the work that must be done on the particle to bring it from a state of rest to a speed v. Thus, the kinetic energy is a measure of the particle’s capacity to do work, which is associated with the motion of the particle. When energy comes from the position of the particle, measured from a fixed datum or reference plane, it is called potential energy. Thus, potential energy is a measure of the amount of work a conservative force will do when it moves from a given position to the datum. In mechanics, the potential energy created by gravity (weight) and an elastic spring is important.

W Gravitational potential energy

Fig. 14–17

Gravitational Potential Energy. If a particle is located a

distance y above an arbitrarily selected datum, as shown in Fig. 14–17, the particle’s weight W has positive gravitational potential energy, Vg , since W has the capacity of doing positive work when the particle is moved back down to the datum. Likewise, if the particle is located a distance y below the datum, Vg is negative since the weight does negative work when the particle is moved back up to the datum. At the datum Vg = 0. In general, if y is positive upward, the gravitational potential energy of the particle of weight W is* Vg = Wy

(14–13)

*Here the weight is assumed to be constant. This assumption is suitable for small differences in elevation y. If the elevation change is significant, however, a variation of weight with elevation must be taken into account (see Prob. 14–81).

Gravitational potential energy of this weight is increased as it is hoisted upward.

14

214

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energy

Elastic Potential Energy.

When an elastic spring is elongated or compressed a distance s from its unstretched position, elastic potential energy Ve can be stored in the spring. This energy is Ve = + 12 ks 2

(14–14)

Here Ve is always positive since, in the deformed position, the force of the spring has the capacity or “potential” for always doing positive work on the particle when the spring is returned to its unstretched position, Fig. 14–18.

14

Unstretched position, s  0 k Ve  0

s k

Ve   12 ks2

s k

Ve   12 ks2 Elastic potential energy

Fig. 14–18

The weight of the sacks resting on this platform causes potential energy to be stored in the supporting springs. As each sack is removed, the platform will rise slightly since some of the potential energy within the springs will be transformed into an increase in gravitational potential energy of the remaining sacks. Such a device is useful for removing the sacks without having to bend over to pick them up as they are unloaded.

14.5

Conservative forCes and potential energy

215

Potential Function. In the general case, if a particle is subjected

to both gravitational and elastic forces, the particle’s potential energy can be expressed as a potential function, which is the algebraic sum V = Vg + Ve

(14–15)

Measurement of V depends on the location of the particle with respect to a selected datum in accordance with Eqs. 14–13 and 14–14. The work done by a conservative force in moving the particle from one point to another point is measured by the difference of this function, i.e., U1 - 2 = V1 - V2

(14–16)

For example, the potential function for a particle of weight W suspended from a spring can be expressed in terms of its position, s, measured from a datum located at the unstretched length of the spring, Fig. 14–19. We have V = Vg + Ve = -Ws + 12 ks2 If the particle moves from s1 to a lower position s2 , then applying Eq. 14–16 it can be seen that the work of W and Fs is U1 - 2 = V1 - V2 =

1 -Ws1

+ 12 ks21 2 -

= W(s2 - s1) -

k

1 -Ws2

1 12 ks22

- 12 ks21 2

Datum s

Fs

W

Fig. 14–19

+ 12 ks22 2

14

216

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energy

When the displacement along the path is infinitesimal, i.e., from point (x, y, z) to (x + dx, y + dy, z + dz), Eq. 14–16 becomes dU = V(x, y, z) - V(x + dx, y + dy, z + dz) = -dV(x, y, z)

(14–17)

If we represent both the force and its displacement as Cartesian vectors, then the work can also be expressed as dU = F # dr = (Fxi + Fyj + Fzk) # (dxi + dyj + dzk)

14

= Fx dx + Fy dy + Fz dz Substituting this result into Eq. 14–17 and expressing the differential dV(x, y, z) in terms of its partial derivatives yields Fx dx + Fy dy + Fz dz = - a

0V 0V 0V dx + dy + dz b 0x 0y 0z

Since changes in x, y, and z are all independent of one another, this equation is satisfied provided Fx = -

0V , 0x

Fy = -

0V , 0y

Fz = -

0V 0z

(14–18)

Thus, F = -

0V 0V 0V i j k 0x 0y 0z

= -a

or

0 0 0 i + j + kbV 0x 0y 0z F = - V

(14–19)

where  (del) represents the vector operator  = (0>0x)i + (0>0y)j + (0>0z)k. Equation 14–19 relates a force F to its potential function V and thereby provides a mathematical criterion for proving that F is conservative. For example, the gravitational potential function for a weight located a distance y above a datum is Vg = Wy. To prove that W is conservative, it is necessary to show that it satisfies Eq. 14–18 (or Eq. 14–19), in which case Fy = -

0V ; 0y

Fy = -

0 (Wy) = -W 0y

The negative sign indicates that W acts downward, opposite to positive y, which is upward.

14.6

14.6

Conservation of energy

217

Conservation of Energy

When a particle is acted upon by a system of both conservative and nonconservative forces, the portion of the work done by the conservative forces can be written in terms of the difference in their potential energies using Eq. 14–16, i.e., (U1 - 2)cons. = V1 - V2 . As a result, the principle of work and energy can be written as T1 + V1 + (U1 - 2)noncons. = T2 + V2

(14–20)

14

Here (U1 - 2)noncons. represents the work of the nonconservative forces acting on the particle. If only conservative forces do work then we have T1 + V1 = T2 + V2

(14–21)

This equation is referred to as the conservation of mechanical energy or simply the conservation of energy. It states that during the motion the sum of the particle’s kinetic and potential energies remains constant. For this to occur, kinetic energy must be transformed into potential energy, and vice versa. For example, if a ball of weight W is dropped from a height h above the ground (datum), Fig. 14–20, the potential energy of the ball is maximum before it is dropped, at which time its kinetic energy is zero. The total mechanical energy of the ball in its initial position is thus E = T1 + V1 = 0 + Wh = Wh When the ball has fallen a distance h>2, its speed can be determined by using v2 = v20 + 2ac(y - y0), which yields v = 22g(h>2) = 2gh. The energy of the ball at the mid-height position is therefore E = T2 + V2 =

1W h 1 2gh 2 2 + Wa b = Wh 2 g 2

Just before the ball strikes the ground, its potential energy is zero and its speed is v = 22gh. Here, again, the total energy of the ball is E = T3 + V3 =

Potential Energy (max) Kinetic Energy (zero)

1W 1 22gh 2 2 + 0 = Wh 2 g

Note that when the ball comes in contact with the ground, it deforms somewhat, and provided the ground is hard enough, the ball will rebound off the surface, reaching a new height h, which will be less than the height  h from which it was first released. Neglecting air friction, the difference in height accounts for an energy loss, El = W(h - h), which occurs during the collision. Portions of this loss produce noise, localized deformation of the ball and ground, and heat.

Potential Energy and Kinetic Energy

h h 2 Datum

Potential Energy (zero) Kinetic Energy (max)

Fig. 14–20

218

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energy

System of Particles.

If a system of particles is subjected only to conservative forces, then an equation similar to Eq. 14–21 can be written for the particles. Applying the ideas of the preceding discussion, Eq. 14–8 (T1 + U1 - 2 = T2) becomes T1 + V1 = T2 + V2

14

(14–22)

Here, the sum of the system’s initial kinetic and potential energies is equal to the sum of the system’s final kinetic and potential energies. In other words, T + V = const.

procedure for analysis The conservation of energy equation can be used to solve problems involving velocity, displacement, and conservative force systems. It is generally easier to apply than the principle of work and energy because this equation requires specifying the particle’s kinetic and potential energies at only two points along the path, rather than determining the work when the particle moves through a displacement. For application it is suggested that the following procedure be used. potential Energy.

• Draw two diagrams showing the particle located at its initial and final points along the path.

• If the particle is subjected to a vertical displacement, establish the

fixed horizontal datum from which to measure the particle’s gravitational potential energy Vg .

• Data pertaining to the elevation y of the particle from the datum

and the stretch or compression s of any connecting springs can be determined from the geometry associated with the two diagrams.

• Recall Vg = Wy, where y is positive upward from the datum and negative downward from the datum; also for a spring, Ve = 12 ks2, which is always positive.

Conservation of Energy. • Apply the equation T1 + V1 = T2 + V2 .

• When determining the kinetic energy, T = 12 mv2, remember that the particle’s speed v must be measured from an inertial reference frame.

14.6

ExamplE

Conservation of energy

14.9

The gantry structure in the photo is used to test the response of an airplane during a crash. As shown in Fig. 14–21a, the plane, having a mass of 8 Mg, is hoisted back until u = 60, and then the pull-back cable AC is released when the plane is at rest. Determine the speed of the plane just before it crashes into the ground, u = 15. Also, what is the maximum tension developed in the supporting cable during the motion? Neglect the size of the airplane and the effect of lift caused by the wings during the motion.

14

Datum C 20 m A u

B

(a) n

Solution Since the force of the cable does no work on the plane, it must be obtained using the equation of motion. First, however, we must determine the plane’s speed at B. potential energy. For convenience, the datum has been established at the top of the gantry, Fig. 14–21a. Conservation of energy.

T

t

15 8000(9.81) N

TA + VA = TB + VB

(b)

0 - 8000 kg (9.81 m>s2)(20 cos 60 m) =

Fig. 14–21

1 2 (8000

kg)v2B - 8000 kg (9.81 m>s2)(20 cos 15 m) vB = 13.52 m>s = 13.5 m>s

Ans.

equation of Motion. From the free-body diagram when the plane is at B, Fig. 14–21b, we have +a

219

Fn = man; T - (8000(9.81) N) cos 15 = (8000 kg) T = 149 kN

(13.52 m>s)2 20 m Ans.

220

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energy

14.10

ExamplE

The ram R shown in Fig. 14–22a has a mass of 100 kg and is released from rest 0.75 m from the top of a spring, A, that has a stiffness kA = 12 kN>m. If a second spring B, having a stiffness k B = 15 kN>m, is “nested” in A, determine the maximum displacement of A needed to stop the downward motion of the ram. The unstretched length of each spring is indicated in the figure. Neglect the mass of the springs.

R

14 0.75 m kA12 kN/m A 0.4 m 0.3 m

B

kB  15 kN/m

Solution potential energy. We will assume that the ram compresses both springs at the instant it comes to rest. The datum is located through the center of gravity of the ram at its initial position, Fig. 14–22b. When the kinetic energy is reduced to zero (v2 = 0), A is compressed a distance sA and B compresses sB = sA - 0.1 m. Conservation of energy.

(a)

T1 + V1 = T2 + V2 0 + 0 = 0 + 0 + 0 = 0 +

5 12 kAs2A

+ 12 kB(sA - 0.1)2 - Wh 6

5 12(12 000 N>m)s2A

+ 12(15 000 N>m)(sA - 0.1 m)2

Rearranging the terms,

981 N

Datum

1

- 981 N (0.75 m + sA) 6

13 500s2A - 2481sA - 660.75 = 0 Using the quadratic formula and solving for the positive root, we have

0.75 m 981 N sA

2

sA = 0.331 m

Ans.

Since sB = 0.331 m - 0.1 m = 0.231 m, which is positive, the assumption that both springs are compressed by the ram is correct.

sA sB  sA  0.1 m (b)

Fig. 14–22

NoTE: The second root, sA = -0.148 m, does not represent the physical situation. Since positive s is measured downward, the negative sign indicates that spring A would have to be “extended” by an amount of 0.148 m to stop the ram.

14.6

ExamplE

221

Conservation of energy

14.11

A smooth 2-kg collar, shown in Fig. 14–23a, fits loosely on the vertical shaft. If the spring is unstretched when the collar is in the position A, determine the speed at which the collar is moving when y = 1 m, if (a) it is released from rest at A, and (b) it is released at A with an upward velocity vA = 2 m>s.

0.75 m A

B k  3 N/m

Solution part (a) potential energy. For convenience, the datum is established through AB, Fig. 14–23b. When the collar is at C, the gravitational potential energy is -(mg)y, since the collar is below the datum, and the elastic potential energy is 12 ks2CB. Here sCB = 0.5 m, which represents the stretch in the spring as shown in the figure.

14

y

C

(a)

Conservation of energy. TA + VA = TC + VC

5 12 ks2CB - mgy 6 5 12(2 kg)v2C 6 + 5 12(3 N>m)(0.5 m)2

0 + 0 = 12 mv2C + 0 + 0 =

vC = 4.39 m>s T

- 2(9.81) N (1 m) 6

(1 m)2  (0.75 m)2  1.25 m W

Ans.

This problem can also be solved by using the equation of motion or the principle of work and energy. Note that for both of these methods the variation of the magnitude and direction of the spring force must be taken into account (see Example 13.4). Here, however, the above solution is clearly advantageous since the calculations depend only on data calculated at the initial and final points of the path.

Datum

0.75 m

A

B

y1m W

vC C

part (b) Conservation of energy. If vA = 2 m>s, using the data in Fig. 14–23b, we have sCB  1.25 m  0.75 m  0.5 m

TA + VA = TC + VC 1 2 2 mvA 1 2 (2

5 12 ks2CB - mgy 6 1 1 2 2 2 (2 kg)vC + 5 2 (3 N>m)(0.5 m)

(b)

+ 0 = 12 mv2C +

kg)(2 m>s)2 + 0 =

vC = 4.82 m>s T

Fig. 14–23

- 2(9.81) N (1 m) 6

Ans.

NoTE: The kinetic energy of the collar depends only on the magnitude

of velocity, and therefore it is immaterial if the collar is moving up or down at 2 m>s when released at A.

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and

energy

preliMiNary prObleMs P14–3. Determine the potential energy of the block that has a weight of 100 N.

14

P14–4. Determine the potential energy in the spring that has an unstretched length of 4 m.

k  10 N/m 5

3m

2m

3

4 4m

Datum (a)

(a) 6m

8m k  10 N/m 5m

3m Datum

4m (b)

(b)

5m k  10 N/m

3m 30 Datum

(c)

Prob. P14–3

(c)

Prob. P14–4

223

Conservation of energy

14.6

FuNDaMeNtal prObleMs F14–13. The 2-kg pendulum bob is released from rest when it is at A. Determine the speed of the bob and the tension in the cord when the bob passes through its lowest position, B. A

F14–16. The 5-kg collar is released from rest at A and travels along the frictionless guide. Determine the speed of the collar when it strikes the stop B. The spring has an unstretched length of 0.5 m. A 1m

14

1.5 m

k  4 N/m 1.5 m B

Prob. F14–13

F14–14. The 2-kg package leaves the conveyor belt at A with a speed of vA = 1 m>s and slides down the smooth ramp. Determine the required speed of the conveyor belt at B so that the package can be delivered without slipping on the belt. Also, find the normal reaction the curved portion of the ramp exerts on the package at B if rB = 2 m. y

vA  1 m/s A

B

Prob. F14–16 F14–17. The 35-kg block is released from rest 1.5 m above the plate. Determine the compression of each spring when the block momentarily comes to rest after striking the plate. Neglect the mass of the plate. The springs are initially unstretched.

4m

1.5 m vB

B

x

Prob. F14–14 F14–15. The 2-kg collar is given a downward velocity of 4 m>s when it is at A. If the spring has an unstretched length of 1 m and a stiffness of k = 30 N>m, determine the velocity of the collar at s = 1 m. 2m

0.075 m A k  16 kN/m

C k  16 kN/m

B

Prob. F14–17

k¿  24 kN/m

F14–18. The 4-kg collar C has a velocity of vA = 2 m>s when it is at A. If the guide rod is smooth, determine the speed of the collar when it is at B. The spring has an unstretched length of k  400 N/m l0 = 0.2 m. A

0.1 m 4 m/s

A

C

B 0.4 m

s k  30 N/m

B

C

Prob. F14–15

Prob. F14–18

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prObleMs 14–66. The assembly consists of two blocks A and B, which have a mass of 20 kg and 30 kg, respectively. Determine the distance B must descend in order for A to achieve a speed of 3 m>s starting from rest.

*14–68. The girl has a mass of 40 kg and center of mass at G. If she is swinging to a maximum height defined by u = 60, determine the force developed along each of the four supporting posts such as AB at the instant u = 0. The swing is centrally located between the posts.

14

A 2m

 G

30

30

B A

B

Prob. 14–68 Prob. 14–66 14–67. The assembly consists of two blocks A and B which have a mass of 20 kg and 30 kg, respectively. Determine the speed of each block when B descends 1.5 m. The blocks are released from rest. Neglect the mass of the pulleys and cords.

14–69. Each of the two elastic rubber bands of the slingshot has an unstretched length of 180 mm. If they are pulled back to the position shown and released from rest, determine the maximum height the 30-g pellet will reach if it is fired vertically upward. Neglect the mass of the rubber bands and the change in elevation of the pellet while it is constrained by the rubber bands. Each rubber band has a stiffness k = 80 N>m. 50 mm 50 mm 240 mm

A

Prob. 14–67

B

Prob. 14–69

14.6 14–70. Two equal-length springs are “nested” together in order to form a shock absorber. If it is designed to arrest the motion of a 2-kg mass that is dropped s = 0.5 m above the top of the springs from an at-rest position, and the maximum compression of the springs is to be 0.2 m, determine the required stiffness of the inner spring, kB, if the outer spring has a stiffness kA = 400 N>m.

225

Conservation of energy

14–73. The roller coaster car has a mass of 700 kg, including its passenger. If it starts from the top of the hill A with a speed vA = 3 m>s, determine the minimum height h of the hill crest so that the car travels around the inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take rB = 7.5 m and rC = 5 m. 14–74. The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, determine the minimum height h of the hill 14 crest so that the car travels around both inside the loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Take rB = 7.5 m and rC = 5 m.

s

A B

Prob. 14–70

A

14–71. The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels down along the smooth guide. Determine the speed of the collar when it reaches point B, which is located just before the end of the curved portion of the rod. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod. *14–72. The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its speed when its center reaches point B and the normal force it exerts on the rod at this point. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod.

B C

h

15 m

10 m

Probs. 14–73/74 14–75. The 2-kg ball of negligible size is fired from point A with an initial velocity of 10 m>s up the smooth inclined plane. Determine the distance from point C to where it hits the horizontal surface at D. Also, what is its velocity when it strikes the surface?

200 mm A B 10 m/s

1.5 m

200 mm k  50 N/m

B

A

C 2m

Probs. 14–71/72

D d

Prob. 14–75

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*14–76. The 4-kg smooth collar has a speed of 3 m>s when it is at s = 0. Determine the maximum distance s it travels before it stops momentarily. The spring has an unstretched length of 1 m.

1.5 m

14

and

energy

14–78. The roller coaster car having a mass m is released from rest at point A. If the track is to be designed so that the car does not leave it at B, determine the required height h. Also, find the speed of the car when it reaches point C. Neglect friction.

A

A

3 m/s

B 7.5 m 20 m

h

s

k  100 N/m

C B

Prob. 14–78 Prob. 14–76 14–77. The spring has a stiffness k = 200 N>m and an unstretched length of 0.5 m. If it is attached to the 3-kg smooth collar and the collar is released from rest at A, determine the speed of the collar when it reaches B. Neglect the size of the collar.

14–79. A 750-mm-long spring is compressed and confined by the plate P, which can slide freely along the vertical 600-mm-long rods.The 40-kg block is given a speed of y = 5 m>s when it is h = 2 m above the plate. Determine how far the plate moves downwards when the block momentarily stops after striking it. Neglect the mass of the plate.

v

A

5 m/s

A

k  200 N/m h

2m

2m

P B k

25 kN/m

600 mm

1.5 m

Prob. 14–77

Prob. 14–79

14.6 *14–80. The spring has a stiffness k = 50 N>m and an unstretched length of 0.3 m. If it is attached to the 2-kg smooth collar and the collar is released from rest at A (u = 0), determine the speed of the collar when u = 60. The motion occurs in the horizontal plane. Neglect the size of the collar.

Conservation of energy

227

14–83. When s = 0, the spring on the firing mechanism is unstretched. If the arm is pulled back such that s = 100 mm and released, determine the speed of the 0.3-kg ball and the normal reaction of the circular track on the ball when u = 60. Assume all surfaces of contact to be smooth. Neglect the mass of the spring and the size of the ball.

z

14

A 2m u

y k  50 N/m

x

u s

Prob. 14–80 14–81. If the mass of the earth is M e, show that the gravitational potential energy of a body of mass m located a distance r from the center of the earth is V g = -GM e m >r. Recall that the gravitational force acting between the earth and the body is F = G(M e m >r2 ), Eq. 13–1. For the calculation, locate the datum at r S  . Also, prove that F is a conservative force. 14–82. A rocket of mass m is fired vertically from the surface of the earth, i.e., at r = r1. Assuming that no mass is lost as it travels upward, determine the work it must do against gravity to reach a distance r2. The force of gravity is F = GM e m >r2 (Eq. 13–1), where M e is the mass of the earth and r the distance between the rocket and the center of the earth.

1.5 m

k  1500 N/m

Prob. 14–83

*14–84. When s = 0, the spring on the firing mechanism is unstretched. If the arm is pulled back such that s = 100 mm and released, determine the maximum angle u the ball will travel without leaving the circular track. Assume all surfaces of contact to be smooth. Neglect the mass of the spring and the size of the ball.

r2

r

r1

u s

1.5 m

k  1500 N/m

Probs. 14–81/82

Prob. 14–84

228

Chapter 14

KinetiCs

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partiCle: WorK

14–85. When the 5-kg box reaches point A it has a speed vA = 10 m>s. Determine the normal force the box exerts on the surface when it reaches point B. Neglect friction and the size of the box.

14

and

energy

14–87. When the 6-kg box reaches point A it has a speed of vA = 2 m>s. Determine the angle u at which it leaves the smooth circular ramp and the distance s to where it falls into the cart. Neglect friction.

y

vA  2 m/s

20 A

yx

B

1.2 m

9m

u x 1/ 2  y1/2  3

B

A

x

s

9m

Prob. 14–87 Prob. 14–85

14–86. When the 5-kg box reaches point A it has a speed vA = 10 m>s. Determine how high the box reaches up the surface before it comes to a stop. Also, what is the resultant normal force on the surface at this point and the acceleration? Neglect friction and the size of the box.

*14–88. The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed vB when he reaches B. Also, compute the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass of 70 kg.

A

y

vB

50 m B

yx

4m

9m s

x 1/ 2  y1/2  3

B

A

x

9m

Prob. 14–86

Prob. 14–88

C 30

14.6 14–89. A 60-kg satellite travels in free flight along an elliptical orbit such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h. What is the speed of the satellite when it reaches point B, where rB = 80 Mm? Hint: See Prob. 14–81, where M e = 5.976(1024) kg and - 12 3 2 # G = 66.73(10 ) m >(kg s ).

229

Conservation of energy

14–91. The 0.75-kg bob of a pendulum is fired from rest at position A by a spring which has a stiffness k = 6 kN>m and is compressed 125 mm. Determine the speed of the bob and the tension in the cord when the bob is at positions B and C. Point B is located on the path where the radius of curvature is still 0.6 m, i.e., just before the cord becomes horizontal.

C

B vB

14 1.2 m

rB  80 Mm 0.6 m rA  20 Mm

vA A

B 0.6 m

Prob. 14–89 A

14–90. The block has a mass of 20 kg and is released from rest when s = 0.5 m. If the mass of the bumpers A and B can be neglected, determine the maximum deformation of each spring due to the collision.

kA  500 N/m

s  0.5 m

A

k

Prob. 14–91

*14–92. The Raptor is an outside loop roller coaster in which riders are belted into seats resembling ski-lift chairs. If the cars travel at v0 = 4 m>s when they are at the top of the hill, determine their speed when they are at the top of the loop and the reaction of the 70-kg passenger on his seat at this instant. The car has a mass of 50 kg. Take h = 12 m, r = 5 m. Neglect friction and the size of the car and passenger.

v0 r

h

B kB  800 N/m

Prob. 14–90

Prob. 14–92

230

Chapter 14

KinetiCs

of a

partiCle: WorK

14–93. If the 20-kg cylinder is released from rest at h = 0, determine the required stiffness k of each spring so that its motion is arrested or stops when h = 0.5 m. Each spring has an unstretched length of 1 m.

2m

14

and

energy

*14–96. The 10-kg sphere C is released from rest when u = 0 and the tension in the spring is 100 N. Determine the speed of the sphere at the instant u = 90. Neglect the mass of rod AB and the size of the sphere.

2m

k

h

k

E

0.4 m

k  500 N/m

Prob. 14–93 14–94. The cylinder has a mass of 20 kg and is released from rest when h = 0. Determine its speed when h = 3 m. Each spring has a stiffness k = 40 N>m and an unstretched length of 2 m.

A

u

D

B

0.3 m

C 0.15 m

2m

2m

Prob. 14–96 k

h

k

Prob. 14–94 14–95. A quarter-circular tube AB of mean radius r contains a smooth chain that has a mass per unit length of m 0. If the chain is released from rest from the position shown, determine its speed when it emerges completely from the tube.

A

O

14–97. A pan of negligible mass is attached to two identical springs of stiffness k = 250 N>m. If a 10-kg box is dropped from a height of 0.5 m above the pan, determine the maximum vertical displacement d. Initially each spring has a tension of 50 N.

1m

1m

k  250 N/m

k  250 N/m

r B

Prob. 14–95

0.5 m d

Prob. 14–97

14.6

Conservation of energy

231

CONCeptual prObleMs C14–1. The roller coaster is momentarily at rest at A. Determine the approximate normal force it exerts on the track at B. Also determine its approximate acceleration at this point. Use numerical data, and take scaled measurements from the photo with a known height at A.

C14–3. The woman pulls the water balloon launcher back, stretching each of the four elastic cords. Estimate the maximum height and the maximum range of a ball placed within the container if it is released from the position shown. Use numerical values and any necessary measurements from the photo. Assume the unstretched length and stiffness of each cord is known. 14

A B

Prob. C14–1 C14–2. As the large ring rotates, the operator can apply a breaking mechanism that binds the cars to the ring, which then allows the cars to rotate with the ring. Assuming the passengers are not belted into the cars, determine the smallest speed of the ring (cars) so that no passenger will fall out. When should the operator release the brake so that the cars can achieve their greatest speed as they slide freely on the ring? Estimate the greatest normal force of the seat on a passenger when this speed is reached. Use numerical values to explain your answer.

Prob. C14–2

Prob. C14–3 C14–4. The girl is momentarily at rest in the position shown. If the unstretched length and stiffness of each of the two elastic cords is known, determine approximately how far the girl descends before she again becomes momentarily at rest. Use numerical values and take any necessary measurements from the photo.

Prob. C14–4

232

Chapter 14

KinetiCs

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partiCle: WorK

and

energy

Chapter revieW F

Work of a Force A force does work when it undergoes a displacement along its line of action. If the force varies with the displacement, then the work is U = 1 F cos u ds. 14

u

s

s1

s2

F cos u

F cos u F cos u

Graphically, this represents the area under the F9s diagram.

s1

s

s2

ds y

If the force is constant, then for a displacement s in the direction of the force, U = Fc s. A typical example of this case is the work of a weight, U = -W y. Here, y is the vertical displacement.

W

s2

s1 s

y2

y1

x

z

Unstretched position, s  0

The work done by a spring force, F = ks, depends upon the stretch or compression s of the spring.

U = - a 12 ks 22 - 12 ks 21 b

s k

Fs Force on Particle

The Principle of Work and Energy If the equation of motion in the tangential direction, Ft = mat, is combined with the kinematic equation, at ds = v dv, we obtain the principle of work and energy. This equation states that the initial kinetic energy T, plus the work done U1 - 2 is equal to the final kinetic energy.

T1 + U192 = T2

Chapter revieW

233

The principle of work and energy is useful for solving problems that involve force, velocity, and displacement. For application, the free-body diagram of the particle should be drawn in order to identify the forces that do work.

14

Power and Efficiency Power is the time rate of doing work. For application, the force F creating the power and its velocity v must be specified. Efficiency represents the ratio of power output to power input. Due to frictional losses, it is always less than one.

P =

dU dt

P = F#v e =

power output power input

Conservation of Energy

Vg  Wy

A conservative force does work that is independent of its path. Two examples are the weight of a particle and the spring force.

W

Friction is a nonconservative force since the work depends upon the length of the path. The longer the path, the more work done. The work done by a conservative force depends upon its position relative to a datum. When this work is referenced from a datum, it is called potential energy. For a weight, it is Vg = { Wy, and for a spring it is Ve = + 12 ks 2. Mechanical energy consists of kinetic energy T and gravitational and elastic potential energies V. According to the conservation of energy, this sum is constant and has the same value at any position on the path. If only gravitational and spring forces cause motion of the particle, then the conservation-of-energy equation can be used to solve problems involving these conservative forces, displacement, and velocity.

y Datum

Vg  0 y

W

Vg  Wy W Gravitational potential energy k

s

Elastic potential energy

T1 + V1 = T2 + V2

Ve  12 ks2

234

Chapter 14

KinetiCs

of a

partiCle: WorK

and

energy

revieW prObleMs R14–1. If a 50-kg crate is released from rest at A, determine its speed after it slides 10 m down the plane. The coefficient of kinetic friction between the crate and plane is mk = 0.3.

R14–3. The block has a mass of 0.75 kg and slides along the smooth chute AB. It is released from rest at A, which has coordinates of A(2.5 m, 0, 5 m). Determine the speed at which it slides off at B, which has coordinates of B(0, 4 m, 0). z

14 2.5 m

A A

10 m

5m B

y

4m

30

x

Prob. R14–1

Prob. R14–3

R14–2. The small 1-kg collar starting from rest at A slides down along the smooth rod. During the motion, the collar is acted upon by a force F = {50i + 30yj + 10zk} N, where x, y, z are in meters. Determine the collar’s speed when it strikes the wall at B.

R14–4. The block has a mass of 0.5 kg and moves within the smooth vertical slot. If the block starts from rest when the attached spring is in the unstretched position at A, determine the constant vertical force F which must be applied to the cord so that the block attains a speed vB = 2.5 m>s when it reaches B; sB = 0.15 m. Neglect the mass of the cord and pulley. 0.3 m

z C 2m A 0.3 m F

B F

sB B

5m

0.5 m

y

A

4m k  100 N/m

x

Prob. R14–2

Prob. R14–4

235

revieW problems R14–5. The crate, having a mass of 25 kg, is hoisted by the pulley system and motor M. If the crate starts from rest and, by constant acceleration, attains a speed of 6 m>s after rising 3 m, determine the power that must be supplied to the motor at the instant s = 3 m. The motor has an efficiency e = 0.74.

R14–7. The collar of negligible size has a mass of 0.25 kg and is attached to a spring having an unstretched length of 100 mm. If the collar is released from rest at A and travels along the smooth guide, determine its speed just before it strikes B.

A M

14 400 mm

200 mm

k  150 N/m

B

Prob. R14–7 s

Prob. R14–5

R14–6. The 25-kg load is hoisted by the pulley system and motor M. If the motor exerts a constant force of 150 N on the cable, determine the power that must be supplied to the motor if the load has been hoisted s = 3 m starting from rest. The motor has an efficiency of e = 0.76.

R14–8. The blocks A and B have the mass of 5 kg and 15 kg, respectively. They are connected together by a light cord and ride in the frictionless grooves. Determine the speed of each block after block A moves 2 m up along the plane. The blocks are released from rest.

z 0.2 m

0.2 m

M

B 5m A

B

s

1m

1m y

x

Prob. R14–6

Prob. R14–8

Chapter 15

(© David J. Green/Alamy) The design of the bumper cars used for this amusement park ride requires knowledge of the principles of impulse and momentum.

Kinetics of a Particle: Impulse and Momentum Chapter ObjeCtives n

To develop the principle of linear impulse and momentum for a particle and apply it to solve problems that involve force, velocity, and time.

n

To study the conservation of linear momentum for particles.

n

To analyze the mechanics of impact.

n

To introduce the concept of angular impulse and momentum.

n

To solve problems involving steady fluid streams and propulsion with variable mass.

15.1

Principle of Linear Impulse and Momentum

In this section we will integrate the equation of motion with respect to time and thereby obtain the principle of impulse and momentum. The resulting equation will be useful for solving problems involving force, velocity, and time. Using kinematics, the equation of motion for a particle of mass m can be written as F = ma = m

dv dt

(15–1)

where a and v are both measured from an inertial frame of reference. Rearranging the terms and integrating between the limits v = v1 at t = t1 and v = v2 at t = t2 , we have 

Lt1

t2

Fdt = m

v2

Lv1

dv

Video Solutions are available for selected questions in this chapter.

238

Chapter 15

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of a

partiCle: impulse

and

momentum

or t2

Fdt = mv2 - mv1 (15–2) Lt1 This equation is referred to as the principle of linear impulse and momentum. From the derivation it can be seen that it is simply a time integration of the equation of motion. It provides a direct means of obtaining the particle’s final velocity v2 after a specified time period when the particle’s initial velocity is known and the forces acting on the particle are either constant or can be expressed as functions of time. By comparison, if v2 was determined using the equation of motion, a two-step process would be necessary; i.e., apply F = ma to obtain a, then integrate a = dv>dt to obtain v2 . 

Linear Momentum. Each of the two vectors of the form L = mv in Eq. 15–2 is referred to as the particle’s linear momentum. Since m is a positive scalar, the linear-momentum vector has the same direction as v, and its magnitude mv has units of mass times velocity, e.g., kg # m>s.

15

The impulse tool is used to remove the dent in the trailer fender. To do so its end is first screwed into a hole drilled in the fender, then the weight is gripped and jerked upwards, striking the stop ring. The impulse developed is transferred along the shaft of the tool and pulls suddenly on the dent.

Linear Impulse. The integral I = 1 F dt in Eq. 15–2 is referred to as

the linear impulse. This term is a vector quantity which measures the effect of a force during the time the force acts. Since time is a positive scalar, the impulse acts in the same direction as the force, and its magnitude has units of force times time, e.g., N # s.* If the force is expressed as a function of time, the impulse can be determined by direct evaluation of the integral. In particular, if the force is constant in both magnitude and direction, the resulting impulse becomes I = 1t12Fc dt = Fc(t2 - t1). t

Graphically the magnitude of the impulse can be represented by the shaded area under the curve of force versus time, Fig. 15–1. A constant force creates the shaded rectangular area shown in Fig. 15–2.

F

F Fc t2

I  t F(t)dt

I  Fc(t2  t1)

1

t1

t2

t

t1

t2

Variable Force

Constant Force

Fig. 15–1

Fig. 15–2

t

*Although the units for impulse and momentum are defined differently, it can be shown that Eq. 15–2 is dimensionally homogeneous.

15.1

prinCiple of linear impulse and momentum

239

Principle of Linear Impulse and Momentum. For problem solving, Eq. 15–2 will be rewritten in the form

mv1 + 

Lt1

t2

F dt = mv2

(15–3)

which states that the initial momentum of the particle at time t1 plus the sum of all the impulses applied to the particle from t1 to t2 is equivalent to the final momentum of the particle at time t2 . These three terms are illustrated graphically on the impulse and momentum diagrams shown in Fig. 15–3. The two momentum diagrams are simply outlined shapes of the particle which indicate the direction and magnitude of the particle’s initial and final momenta, mv1 and mv2. Similar to the free-body diagram, the impulse diagram is an outlined shape of the particle showing all the impulses that act on the particle when it is located at some intermediate point along its path. If each of the vectors in Eq. 15–3 is resolved into its x, y, z components, we can write the following three scalar equations of linear impulse and momentum.

m(vx )1 +  m(vy )1 +  m(vz )1 + 



Lt1

t2

Lt1

Fx dt = m(vx )2 t2

Lt1

t2 t1

Fy dt = m(vy )2

(15–4)

t2

Fz dt = m(vz )2

F dt

mv1

+ Initial momentum diagram

= Impulse diagram

Fig. 15–3

mv2 Final momentum diagram

15

The study of many types of sports, such as golf, requires application of the principle of linear impulse and momentum.

240

Chapter 15

KinetiCs

of a

partiCle: impulse

15.2

Fi i

G rG

Principle of Linear Impulse and Momentum for a System of Particles

Fi = m i

ri y

Inertial coordinate system

15

momentum

The principle of linear impulse and momentum for a system of particles moving relative to an inertial reference, Fig. 15–4, is obtained from the equation of motion applied to all the particles in the system, i.e.,

z

fi

and

x

Fig. 15–4

dvi dt

(15–5)

The term on the left side represents only the sum of the external forces acting on the particles. Recall that the internal forces fi acting between particles do not appear with this summation, since by Newton’s third law they occur in equal but opposite collinear pairs and therefore cancel out. Multiplying both sides of Eq. 15–5 by dt and integrating between the limits t = t1, vi = (vi)1 and t = t2, vi = (vi)2 yields

m i (vi)1 + 

Lt1

t2

Fi dt = m i (vi)2

(15–6)

This equation states that the initial linear momentum of the system plus the impulses of all the external forces acting on the system from t1 to t2 is equal to the system’s final linear momentum. Since the location of the mass center G of the system is determined from mrG = m iri , where m = m i is the total mass of all the particles, Fig. 15–4, then taking the time derivative, we have mvG = m ivi which states that the total linear momentum of the system of particles is equivalent to the linear momentum of a “fictitious” aggregate particle of mass m = m i moving with the velocity of the mass center of the system. Substituting into Eq. 15–6 yields

m(vG)1 + 

Lt1

t2

Fi dt = m(vG)2

(15–7)

Here the initial linear momentum of the aggregate particle plus the external impulses acting on the system of particles from t1 to t2 is equal to the aggregate particle’s final linear momentum. As a result, the above equation justifies application of the principle of linear impulse and momentum to a system of particles that compose a rigid body.

15.2

prinCiple of linear impulse and momentum for a system of partiCles

241

procedure for analysis The principle of linear impulse and momentum is used to solve problems involving force, time, and velocity, since these terms are involved in the formulation. For application it is suggested that the following procedure be used.* Free-Body Diagram. • Establish the x, y, z inertial frame of reference and draw the particle’s free-body diagram in order to account for all the forces that produce impulses on the particle.

• The direction and sense of the particle’s initial and final velocities should be established.

• If a vector is unknown, assume that the sense of its components is in the direction of the positive inertial coordinate(s).

• As an alternative procedure, draw the impulse and momentum diagrams for the particle as discussed in reference to Fig. 15–3.

Principle of Impulse and Momentum. apply the • In accordance with the established coordinate system, t principle of linear impulse and momentum, mv1 +  1t12F dt = mv2 . If motion occurs in the x–y plane, the two scalar component equations can be formulated by either resolving the vector components of F from the free-body diagram, or by using the data on the impulse and momentum diagrams.

• Realize that every force acting on the particle’s free-body diagram will create an impulse, even though some of these forces will do no work.

• Forces that are functions of time must be integrated to obtain the

impulse. Graphically, the impulse is equal to the area under the force–time curve.

Ndt Fdt F¿dt Wt

N¿dt

As the wheels of the pitching machine rotate, they apply frictional impulses to the ball, thereby giving it a linear momentum. These impulses are shown on the impulse diagram. Here both the frictional and normal impulses vary with time. By comparison, the weight impulse is constant and is very small since the time t the ball is in contact with the wheels is very small.

*This procedure will be followed when developing the proofs and theory in the text.

15

242

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ExaMPlE

KinetiCs

of a

partiCle: impulse

and

momentum

15.1 200 N

45

The 100-kg crate shown in Fig. 15–5a is originally at rest on the smooth horizontal surface. If a towing force of 200 N, acting at an angle of 45°, is applied for 10 s, determine the final velocity and the normal force which the surface exerts on the crate during this time interval. SolutIon This problem can be solved using the principle of impulse and momentum since it involves force, velocity, and time.

(a)

Free-body Diagram. See Fig. 15–5b. Since all the forces acting are constant, the impulses are simply the product of the force magnitude and 10 s [I = Fc(t2 - t1)]. Note the alternative procedure of drawing the crate’s impulse and momentum diagrams, Fig. 15–5c.

15

principle of impulse and Momentum. Applying Eqs. 15–4 yields

y v x

200 N

981 N

+ ) (S

m(vx )1 + 

Lt1

t2

Fx dt = m(vx )2

0 + 200 N cos 45(10 s) = (100 kg)v2

45

v2 = 14.1 m>s NC

(+c)

m(vy )1 + 

(b)

Lt1

Ans.

t2

Fy dt = m(vy )2

0 + N C(10 s) - 981 N(10 s) + 200 N sin 45(10 s) = 0 Ans.

N C = 840 N

NOTE: Since no motion occurs in the y direction, direct application of the equilibrium equation Fy = 0 gives the same result for N C . Try to solve the problem by first applying Fx = max, then v = v0 + ac t. 200 N (10 s)

981 N (10 s)

45

+ NC (10 s) (c)

Fig.15–5

=

(100 kg) v2

15.2

ExaMPlE

prinCiple of linear impulse and momentum for a system of partiCles

15.2

The 25-kg crate shown in Fig. 15–6a is acted upon by a force having a variable magnitude P = (100t) N, where t is in seconds. Determine the crate’s velocity 2 s after P has been applied. The initial velocity is v1 = 1 m>s down the plane, and the coefficient of kinetic friction between the crate and the plane is mk = 0.3. SolutIon Free-body Diagram. See Fig. 15–6b. Since the magnitude of force P = 100t varies with time, the impulse it creates must be determined by integrating over the 2-s time interval. principle of impulse and Momentum. x direction, we have ( +b)

m(vx )1 +  2s

Lt1

v1  1 m/s

P

30 (a)

Applying Eqs. 15–4 in the

15

t2

Fx dt = m(vx )2

L0 25 + 200 - 0.6NC + 245.25 = 25v2

(25 kg) (1 m>s) +

2

100t dt - 0.3NC (2 s) + (25 kg)(9.81 m>s ) sin 30(2 s) = (25 kg) v2

The equation of equilibrium can be applied in the y direction. Why? + aFy = 0;

25(9.81)N y

NC - 25(9.81) cos 30 N = 0

Solving,

30

v

NC = 212.39 N v2 = 13.7 m>s b

x

F  0.3 NC

P  100t

NC

Ans.

NOTE: We can also solve this problem using the equation of motion.

From Fig. 15–6b,

(b)

Fig. 15–6

+ bFx = max ; 100t - 0.3(212.39) + 25(9.81) sin 30 = 25 a a = 4t + 2.356 Using kinematics v

+ bdv = a dt;

243

L1 m>s

dv =

L0

2s

(4t + 2.356)dt

v = 13.7 m>s

Ans.

By comparison, application of the principle of impulse and momentum eliminates the need for using kinematics (a = dv>dt) and thereby yields an easier method for solution.

244

Chapter 15

KinetiCs

partiCle: impulse

and

momentum

15.3

ExaMPlE

C

Datum

sA D sB A 3 kg

Blocks A and B shown in Fig. 15–7a have a mass of 3 kg and 5 kg, respectively. If the system is released from rest, determine the velocity of block B in 6 s. Neglect the mass of the pulleys and cord. SolutIon Free-body Diagram. See Fig. 15–7b. Since the weight of each block is constant, the cord tensions will also be constant. Furthermore, since the mass of pulley D is neglected, the cord tension T A = 2T B . Note that the blocks are both assumed to be moving downward in the positive coordinate directions, sA and sB . principle of impulse and Momentum. Block A:

B

15

of a

5 kg (a)

(+T)

m(vA )1 + 

Lt1

t2

Fy dt = m(vA )2

0 - 2T B (6 s) + 3(9.81) N(6 s) = (3 kg)(vA )2

TB TB

(1)

Block B: t2

(+ T ) TA  2TB 2TB

vA sA 3(9.81) N

TB

Fy dt = m(vB )2 Lt1 0 + 5(9.81) N(6 s) - T B (6 s) = (5 kg)(vB )2 m(vB )1 + 

(2)

Kinematics. Since the blocks are subjected to dependent motion, the velocity of A can be related to that of B by using the kinematic analysis discussed in Sec. 12.9. A horizontal datum is established through the fixed point at C, Fig. 15–7a, and the position coordinates, sA and sB , are related to the constant total length l of the vertical segments of the cord by the equation 2sA + sB = l Taking the time derivative yields 2vA = -vB

(3)

As indicated by the negative sign, when B moves downward A moves upward. Substituting this result into Eq. 1 and solving Eqs. 1 and 2 yields

vB

(vB )2 = 35.8 m>s T sB 5(9.81) N (b)

Fig. 15–7

Ans.

T B = 19.2 N NOTE: Realize that the positive (downward) direction for vA and vB is

consistent in Figs. 15–7a and 15–7b and in Eqs. 1 to 3. This is important since we are seeking a simultaneous solution of equations.

15.2

prinCiple of linear impulse and momentum for a system of partiCles

245

preLiMiNarY prObLeMs 15–1.

Determine the impulse of the force for t = 2 s.

e)

k  10 N/m 80 N

a) f) 100 N

15 30

60 N

3

5 4

Prob. P15–1

b)

200 N

15–2.

Determine the linear momentum of the 10-kg

block. a)

10 m/s

6m

c) F  (6t) N 3

5

b)

4

2 m/s

30

d)

c)

F (N)

60 N 3 m/s

F 30

100 N

20

1

3

t (s)

Prob. P15–2

45

246

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

FuNDaMeNtaL prObLeMs F15–1. The 0.5-kg ball strikes the rough ground and rebounds with the velocities shown. Determine the magnitude of the impulse the ground exerts on the ball. Assume that the ball does not slip when it strikes the ground, and neglect the size of the ball and the impulse produced by the weight of the ball.

F15–4. The wheels of the 1.5-Mg car generate the traction force F described by the graph. If the car starts from rest, determine its speed when t = 6 s. F (kN)

F 6 kN v2  10 m/s

v1  25 m/s

15 45

2

30

6

t (s)

Prob. F15–4

Prob. F15–1

F15–2. If the coefficient of kinetic friction between the 75-kg crate and the ground is mk = 0.2, determine the speed of the crate when t = 4 s. The crate starts from rest and is towed by the 500-N force.

F15–5. The 2.5-Mg four-wheel-drive SUV tows the 1.5-Mg trailer. The traction force developed at the wheels is FD = 9 kN. Determine the speed of the truck in 20 s, starting from rest. Also, determine the tension developed in the coupling, A, between the SUV and the trailer. Neglect the mass of the wheels.

A 500 N

FD

30

Prob. F15–5 Prob. F15–2

F15–3. The motor exerts a force of F = (20t2) N on the cable, where t is in seconds. Determine the speed of the 25-kg crate when t = 4 s. The coefficients of static and kinetic friction between the crate and the plane are ms = 0.3 and mk = 0.25, respectively.

F15–6. The 10-kg block A attains a velocity of 1 m>s in 5 seconds, starting from rest. Determine the tension in the cord and the coefficient of kinetic friction between block A and the horizontal plane. Neglect the weight of the pulley. Block B has a mass of 8 kg. A

A B

Prob. F15–3

Prob. F15–6

15.2

247

prinCiple of linear impulse and momentum for a system of partiCles

prObLeMs 15–1. A man kicks the 150-g ball such that it leaves the ground at an angle of 60 and strikes the ground at the same elevation a distance of 12 m away. Determine the impulse of his foot on the ball at A. Neglect the impulse caused by the ball’s weight while it’s being kicked.

*15–4. A train consists of a 50-Mg engine and three cars, each having a mass of 30 Mg. If it takes 80 s for the train to increase its speed uniformly to 40 km>h, starting from rest, determine the force T developed at the coupling between the engine E and the first car A. The wheels of the engine provide a resultant frictional tractive force F which gives the train forward motion, whereas the car wheels roll freely. Also, determine F acting on the engine wheels.

v

15 v A

60

E F

A

Prob. 15–4 Prob. 15–1 15–2. A 2.5-kg block is given an initial velocity of 3 m>s up a 45° smooth slope. Determine the time for it to travel up the slope before it stops. 15–3. A hockey puck is traveling to the left with a velocity of v1 = 10 m>s when it is struck by a hockey stick and given a velocity of v2 = 20 m>s as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of 0.2 kg.

15–5. The winch delivers a horizontal towing force F to its cable at A which varies as shown in the graph. Determine the speed of the 70-kg bucket when t = 18 s. Originally the bucket is moving upward at v1 = 3 m>s. 15–6. The winch delivers a horizontal towing force F to its cable at A which varies as shown in the graph. Determine the speed of the 80-kg bucket when t = 24 s. Originally the bucket is released from rest.

A F

F (N) 600

v2  20 m/s 40 v1  10 m/s

Prob. 15–3

360

v B

12

Probs. 15–5/6

24

t (s)

248

Chapter 15

KinetiCs

of a

partiCle: impulse

15–7. The 50-kg crate is pulled by the constant force P. If the crate starts from rest and achieves a speed of 10 m>s in 5 s, determine the magnitude of P. The coefficient of kinetic friction between the crate and the ground is mk = 0.2.

P

and

momentum

15–10. The 200-kg crate rests on the ground for which the coefficients of static and kinetic friction are μs = 0.5 and μk = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the crate when t = 4 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the crate. T (N)

30

800 T  400t1/2

Prob. 15–7

t (s)

4

15

A T

*15–8. If the jets exert a vertical thrust of T = (500t3>2) N, where t is in seconds, determine the man’s speed when t = 3 s. The total mass of the man and the jet suit is 100 kg. Neglect the loss of mass due to the fuel consumed during the lift which begins from rest on the ground.

Prob. 15–10 15–11. The 2.5-Mg van is traveling with a speed of 100 km>h when the brakes are applied and all four wheels lock. If the speed decreases to 40 km>h in 5 s, determine the coefficient of kinetic friction between the tires and the road.

T

Prob. 15–11 Prob. 15–8

15–9. The train consists of a 30-Mg engine E, and cars A, B, and C, which have a mass of 15 Mg, 10 Mg, and 8 Mg, respectively. If the tracks provide a traction force of F = 30 kN on the engine wheels, determine the speed of the train when t  =  30  s, starting from rest. Also, find the horizontal coupling force at D between the engine E and car A. Neglect rolling resistance.

C

B

A

*15–12. During operation the jack hammer strikes the concrete surface with a force which is indicated in the graph. To achieve this the 2-kg spike S is fired into the surface at 90 m>s. Determine the speed of the spike just after rebounding.

F (kN) 1500

E D

Probs. 15–9

F

S 30 kN

0 0

0.1

0.4

Prob. 15–12

t (ms)

15.2

249

prinCiple of linear impulse and momentum for a system of partiCles

15–13. For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is FD = (600t2) N, where t is in seconds. If the van has a speed of 20 km>h when t = 0, determine its speed when t = 5 s.

*15–16. Under a constant thrust of T = 40 kN, the 1.5-Mg dragster reaches its maximum speed of 125 m>s in 8 s starting from rest. Determine the average drag resistance FD during this period of time.

FD

Prob. 15–13 15–14. The motor, M, pulls on the cable with a force F = (10t2 + 300) N, where t is in seconds. If the 100 kg crate is originally at rest at t = 0, determine its speed when t = 4 s. Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate.

T

FD

40 kN

15 Prob. 15–16

M

15–17. The thrust on the 4-Mg rocket sled is shown in the graph. Determine the sleds maximum velocity and the distance the sled travels when t = 35 s. Neglect friction.

Prob. 15–14 15–15. A tankcar has a mass of 20 Mg and is freely rolling to the right with a speed of 0.75 m>s. If it strikes the barrier, determine the horizontal impulse needed to stop the car if the spring in the bumper B has a stiffness (a) k S ∞ (bumper is rigid), and (b) k = 15 kN>m.

T T (kN) 20

v  0.75 m/s

T  4 t1/2 k B 25

Prob. 15–15

Prob. 15–17

35

t (s)

250

Chapter 15

KinetiCs

of a

partiCle: impulse

15–18. A 50-kg crate rests against a stop block s, which prevents the crate from moving down the plane. If the coefficients of static and kinetic friction between the plane and the crate are ms = 0.3 and mk = 0.2, respectively, determine the time needed for the force F to give the crate a speed of 2 m>s up the plane.The force always acts parallel to the plane and has a magnitude of F = (300t) N, where t is in seconds. Hint: First determine the time needed to overcome static friction and start the crate moving.

and

momentum

*15–20. The choice of a seating material for moving vehicles depends upon its ability to resist shock and vibration. From the data shown in the graphs, determine the impulses created by a falling weight onto a sample of urethane foam and CONFOR foam.

F (N) 1.2

F

urethane 0.8

15

CONFOR

0.5 0.4 0.3

s 30

t (ms) 2

Prob. 15–18

4

7

10

14

Prob. 15–20 15–19. The towing force acting on the 400-kg safe varies as shown on the graph. Determine its speed, starting from rest, when t = 8 s. How far has it traveled during this time?

F

15–21. If it takes 35 s for the 50-Mg tugboat to increase its speed uniformly to 25 km>h, starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine F acting on the tugboat. The barge has a mass of 75 Mg.

F (N) 750 600

5

8

Prob. 15–19

t (s)

F Prob. 15–21

15.2

251

prinCiple of linear impulse and momentum for a system of partiCles

15–22. The crate B and cylinder A have a mass of 200 kg and 75 kg, respectively. If the system is released from rest, determine the speed of the crate and cylinder when t = 3 s. Neglect the mass of the pulleys.

*15–24. The 30-kg slider block is moving to the left with a speed of 5 m>s when it is acted upon by the forces F1 and F2. If these loadings vary in the manner shown on the graph, determine the speed of the block at t = 6 s. Neglect friction and the mass of the pulleys and cords.

F2

F1

F (N)

B

15

F2

40 30 20

A

F1

10

Prob. 15–22

0

2

4

6

t (s)

Prob. 15–24 15–23. The motor exerts a force F on the 40-kg crate as shown in the graph. Determine the speed of the crate when t = 3 s and when t = 6 s. When t = 0, the crate is moving downward at 10 m>s.

15–25. The balloon has a total mass of 400 kg including the passengers and ballast. The balloon is rising at a constant velocity of 18 km>h when h = 10 m. If the man drops the 40-kg sand bag, determine the velocity of the balloon when the bag strikes the ground. Neglect air resistance.

A F vA  18 km/h F (N) 450

B

A

150 6

Prob. 15–23

t (s)

h

Prob. 15–25

252

Chapter 15

KinetiCs

of a

partiCle: impulse

15–26. As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the 10-kg block which slides along the smooth surface and is subjected to a horizontal force of 6 N. If observer A is in a fixed frame x, determine the final speed of the block in 4 s if it has an initial speed of 5 m>s measured from the fixed frame. Compare the result with that obtained by an observer B, attached to the x  axis that moves at a constant velocity of 2 m>s relative to A.

A

x

B

x¿

and

momentum

15–29. In case of emergency, the gas actuator is used to move a 75-kg block B by exploding a charge C near a pressurized cylinder of negligible mass. As a result of the explosion, the cylinder fractures and the released gas forces the front part of the cylinder, A, to move B forward, giving it a speed of 200 mm>s in 0.4 s. If the coefficient of kinetic friction between B and the floor is μk = 0.5, determine the impulse that the actuator imparts to B.

B

2 m/s

15

A C

5 m/s

vB  200 mm/s

6N

B

Prob. 15–26

A

15–27. The 20-kg crate is lifted by a force of F = (100 + 5t2) N, where t is in seconds. Determine the speed of the crate when t = 3 s, starting from rest. *15–28. The 20-kg crate is lifted by a force of F =  (100 + 5t2) N, where t is in seconds. Determine how high the crate has moved upward when t = 3 s, starting from rest.

F

Prob. 15–29

15–30. A jet plane having a mass of 7 Mg takes off from an aircraft carrier such that the engine thrust varies as shown by the graph. If the carrier is traveling forward with a speed of 40 km>h, determine the plane’s airspeed after 5 s.

F (kN) 40 km/h

B

15

5 A 0

Probs. 15–27/28

2

5

Prob. 15–30

t (s)

15.2

15–31. The 6-kg block is moving downward at y1 = 3 m>s when it is 8 m from the sandy surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

v1

253

prinCiple of linear impulse and momentum for a system of partiCles

15–33. The log has a mass of 500 kg and rests on the ground for which the coefficients of static and kinetic friction are ms = 0.5 and mk = 0.4, respectively. The winch delivers a horizontal towing force T to its cable at A which varies as shown in the graph. Determine the speed of the log when t = 5 s. Originally the tension in the cable is zero. Hint: First determine the force needed to begin moving the log.

3 m/s T (N) 1800

8m

T  200 t2

15 3

A T

Prob. 15–31 *15–32. The 6-kg block is falling downward at y1 = 3 m>s when it is 8 m from the sandy surface. Determine the average impulsive force acting on the block by the sand if the motion of the block is stopped in time 1.2 s once the block strikes the sand. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

v1

t (s)

Prob. 15–33

15–34. The 0.15-kg baseball has a speed of v = 30 m>s just before it is struck by the bat. It then travels along the trajectory shown before the outfielder catches it. Determine the magnitude of the average impulsive force imparted to the ball if it is in contact with the bat for 0.75 ms.

3 m/s v2 15 v1  30 m/s 15 0.75 m

8m

100 m

Prob. 15–32

Prob. 15–34

2.5 m

254

Chapter 15

KinetiCs

of a

partiCle: impulse

15.3 Please refer to the Companion Website for the animation: Impact of 2 Sliding Masses

and

momentum

Conservation of Linear Momentum for a System of Particles

When the sum of the external impulses acting on a system of particles is zero, Eq. 15–6 reduces to a simplified form, namely, m i (vi)1 = m i (vi)2

(15–8)

This equation is referred to as the conservation of linear momentum. It states that the total linear momentum for a system of particles remains constant during the time period t1 to t2 . Substituting mvG = m ivi into Eq. 15–8, we can also write

15

(vG)1 = (vG)2

The hammer in the top photo applies an impulsive force to the stake. During this extremely short time of contact the weight of the stake  can be considered nonimpulsive, and provided the stake is driven into soft ground, the impulse of the ground acting on the stake can  also be considered nonimpulsive. By contrast, if the stake is used in a concrete chipper to break concrete, then two impulsive forces act on the stake: one at its top due to the chipper and the other on its bottom due to the rigidity of the concrete.

(15–9)

which indicates that the velocity vG of the mass center for the system of particles does not change if no external impulses are applied to the system. The conservation of linear momentum is often applied when particles collide or interact. For application, a careful study of the free-body diagram for the entire system of particles should be made in order to identify the forces which create either external or internal impulses and thereby determine in what direction(s) linear momentum is conserved. As stated earlier, the internal impulses for the system will always cancel out, since they occur in equal but opposite collinear pairs. If the time period over which the motion is studied is very short, some of the external impulses may also be neglected or considered approximately equal to zero. The forces causing these negligible impulses are called nonimpulsive forces. By comparison, forces which are very large and act for a very short period of time produce a significant change in momentum and are called impulsive forces. They, of course, cannot be neglected in the impulse–momentum analysis. Impulsive forces normally occur due to an explosion or the striking of one body against another, whereas nonimpulsive forces may include the weight of a body, the force imparted by a slightly deformed spring having a relatively small stiffness, or for that matter, any force that is very small compared to other larger (impulsive) forces. When making this distinction between impulsive and nonimpulsive forces, it is important to realize that this only applies during the time t1 to t2 . To illustrate, consider the effect of striking a tennis ball with a racket as shown in the photo. During the very short time of interaction, the force of the racket on the ball is  impulsive since it changes the ball’s momentum drastically. By comparison, the ball’s weight will have a negligible effect on the change

15.3

Conservation of linear momentum for a system of partiCles

255

in momentum, and therefore it is nonimpulsive. Consequently, it can be neglected from an impulse–momentum analysis during this time. If an impulse–momentum analysis is considered during the much longer time of flight after the racket–ball interaction, then the impulse of the ball’s weight is important since it, along with air resistance, causes the change in the momentum of the ball.

procedure for analysis Generally, the principle of linear impulse and momentum or the conservation of linear momentum is applied to a system of particles in order to determine the final velocities of the particles just after the time period considered. By applying this principle to the entire system, the internal impulses acting within the system, which may be unknown, are eliminated from the analysis. For application it is suggested that the following procedure be used. Free-Body Diagram.

• Establish the x, y, z inertial frame of reference and draw the free-

body diagram for each particle of the system in order to identify the internal and external forces.

• The conservation of linear momentum applies to the system in a

direction which either has no external forces or the forces can be considered nonimpulsive.

• Establish the direction and sense of the particles’ initial and final

velocities. If the sense is unknown, assume it is along a positive inertial coordinate axis.

• As an alternative procedure, draw the impulse and momentum diagrams for each particle of the system.

Momentum Equations. • Apply the principle of linear impulse and momentum or the conservation of linear momentum in the appropriate directions.

• If it is necessary to determine the internal impulse 1 F dt acting on only one particle of a system, then the particle must be isolated (free-body diagram), and the principle of linear impulse and momentum must be applied to this particle.

• After the impulse is calculated, and provided the time t for which the impulse acts is known, then the average impulsive force Favg can be determined from Favg = 1 F dt> t.

15

256

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

15.4

ExaMPlE

The 15-Mg boxcar A is coasting at 1.5 m>s on the horizontal track when it encounters a 12-Mg tank car B coasting at 0.75 m>s toward it as shown in Fig. 15–8a. If the cars collide and couple together, determine (a) the speed of both cars just after the coupling, and (b) the average force between them if the coupling takes place in 0.8 s. 1.5 m/s

0.75 m/s

A

B

(a)

15

v x F

F (b)

SolutIon part (a) Free-body Diagram.* Here we have considered both cars as a single system, Fig. 15–8b. By inspection, momentum is conserved in the x direction since the coupling force F is internal to the system and will therefore cancel out. It is assumed both cars, when coupled, move at v2 in the positive x direction. Conservation of Linear Momentum. + ) (S

m A (vA )1 + m B (vB )1 = (m A + m B )v2 (15 000 kg)(1.5 m>s) - 12 000 kg(0.75 m>s) = (27 000 kg)v2 v2 = 0.5 m>s S

part (b). The average (impulsive) coupling force, Favg , can be determined by applying the principle of linear momentum to either one of the cars.

v x F (c)

Fig. 15–8

Ans.

Free-body Diagram. As shown in Fig. 15–8c, by isolating the boxcar the coupling force is external to the car. principle of impulse and Momentum. = Favg(0.8 s), we have + ) (S

Since 1 F dt = Favg t

F dt = m A v2 L (15 000 kg)(1.5 m>s) - Favg(0.8 s) = (15 000 kg)(0.5 m>s) m A (vA )1 + 

Favg = 18.8 kN

Ans.

NOTE: Solution was possible here since the boxcar’s final velocity was obtained in Part (a). Try solving for Favg by applying the principle of impulse and momentum to the tank car. *Only horizontal forces are shown on the free-body diagram.

15.3

ExaMPlE

15.5

The bumper cars A and B in Fig. 15–9a each have a mass of 150 kg and are coasting with the velocities shown before they freely collide head on. If no energy is lost during the collision, determine their velocities after collision.

(vA)1  3 m/s (vB)1  2 m/s A

B

SolutIon Free-body Diagram. The cars will be considered as a single system. The free-body diagram is shown in Fig. 15–9b. Conservation of Momentum. + ) (S

257

Conservation of linear momentum for a system of partiCles

m A (vA)1 + m B(vB)1 = m A(vA)2 + m B(vB)2

(150 kg)(3 m>s) + (150 kg)( -2 m>s) = (150 kg)(vA)2 + (150 kg)(vB)2

(vA)2 = 1 - (vB)2

(1)

Conservation of energy. Since no energy is lost, the conservation of energy theorem gives T 1 + V 1 = T2 + V2 1 1 1 1 m (v )2 + m B (vB )12 + 0 = m A (vA )22 + m B (vB )22 + 0 2 A A1 2 2 2 1 1 1 (150 kg)(3 m>s)2 + (150 kg)(2 m>s)2 + 0 = (150 kg)(vA)22 2 2 2 1 + (150 kg)(vB )22 + 0 2 (vA )22 + (vB )22 = 13 (2) Substituting Eq. (1) into (2) and simplifying, we get (vB )22 - (vB )2 - 6 = 0 Solving for the two roots, (vB)2 = 3 m>s

and

(vB)2 = -2 m>s

Since (vB)2 = -2 m>s refers to the velocity of B just before collision, then the velocity of B just after the collision must be (vB)2 = 3 m>s S Ans. Substituting this result into Eq. (1), we obtain (vA)2 = 1 - 3 m>s = -2 m>s = 2 m>s d

Ans.

(a) 150(9.81) N 150(9.81) N

A

B F

F NB

NA (b)

Fig. 15–9

15

258

Chapter 15

ExaMPlE

KinetiCs

of a

partiCle: impulse

and

momentum

15.6 An 800-kg rigid pile shown in Fig. 15–10a is driven into the ground using a 300-kg hammer. The hammer falls from rest at a height y 0 = 0.5 m and strikes the top of the pile. Determine the impulse which the pile exerts on the hammer if the pile is surrounded entirely by loose sand so that after striking, the hammer does not rebound off the pile. SolutIon Conservation of energy. The velocity at which the hammer strikes the pile can be determined using the conservation of energy equation applied to the hammer. With the datum at the top of the pile, Fig. 15–10a, we have T 0 + V 0 = T1 + V1

15

H

1 1 m H (vH) 20 + WHy0 = mH (vH) 21 + WHy1 2 2 1 0 + 300(9.81) N(0.5 m) = (300 kg)(vH)21 + 0 2 (vH)1 = 3.132 m>s

y0  0.5 m Datum

P

Free-body Diagram. From the physical aspects of the problem, the free-body diagram of the hammer and pile, Fig. 15–10b, indicates that during the short time from just before to just after the collision, the weights of the hammer and pile and the resistance force Fs of the sand are all nonimpulsive. The impulsive force R is internal to the system and therefore cancels. Consequently, momentum is conserved in the vertical direction during this short time. Conservation of Momentum. Since the hammer does not rebound off the pile just after collision, then (vH)2 = (vP)2 = v2 .

Sand

(a) WH  0 R R

v

(+ T )

WP  0 y

principle of impulse and Momentum. The impulse which the pile imparts to the hammer can now be determined since v2 is known. From the free-body diagram for the hammer, Fig. 15–10c, we have

Fs  0

t2

(+ T )

(b) WH  0 v R

m H (vH)1 + mP (vP)1 = mHv2 + mP v2 (300 kg)(3.132 m>s) + 0 = (300 kg)v2 + (800 kg)v2 v2 = 0.8542 m>s

y (c)

Fig. 15–10

m H (vH)1 + 

(300 kg)(3.132 m>s) -

L

L

Lt1

Fy dt = mH v2

R dt = (300 kg)(0.8542 m>s) R dt = 683 N # s

Ans.

NOTE: The equal but opposite impulse acts on the pile. Try finding this impulse by applying the principle of impulse and momentum to the pile.

15.3

ExaMPlE

Conservation of linear momentum for a system of partiCles

259

15.7

The 80-kg man can throw the 20-kg box horizontally at 4 m>s when standing on the ground. If instead he firmly stands in the 120-kg boat and throws the box, as shown in the photo, determine how far the boat will move in three seconds. Neglect water resistance. SolutIon Free-body Diagram. If the man, boat, and box are considered as a single system, the horizontal forces between the man and the boat and the man and the box become internal to the system, Fig. 15–11a, and so linear momentum will be conserved along the x axis. Conservation of Momentum. When writing the conservation of momentum equation, it is important that the velocities be measured from the same inertial coordinate system, assumed here to be fixed. From this coordinate system, we will assume that the boat and man go vbox to the right while the box goes to the left, as shown in Fig. 15–11b. Applying the conservation of linear momentum to the man, boat, box system, + ) (S

F

F F

F

15 (a)

vb

0 + 0 + 0 = (mm + mb) vb − mbox vbox 0 = (80 kg + 120 kg) vb − (20 kg) vbox vbox = 10 vb

(b)

(1)

Kinematics. Since the velocity of the box relative to the man (and boat), vbox>b, is known, then vb can also be related to vbox using the relative velocity equation. + ) (S

vbox = vb + vbox>b

Solving Eqs. (1) and (2),

−vbox = vb − 4 m>s

(2)

vbox = 3.64 m>s d

vb = 0.3636 m>s S

The displacement of the boat in three seconds is therefore sb = vbt = (0.3636 m>s)(3 s) = 1.09 m

Ans.

Fig. 15–11

260

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

15.8

ExaMPlE

The 600-kg cannon shown in Fig. 15–12a fires an 4-kg projectile with a muzzle velocity of 450 m>s measured relative to the cannon. If firing takes place in 0.03 s, determine the recoil velocity of the cannon just after firing. The cannon support is fixed to the ground, and the horizontal recoil of the cannon is absorbed by two springs.

Recoil spring

(a)

SolutIon part (a) Free-body Diagram.* As shown in Fig. 15–12b, we will consider the projectile and cannon as a single system, since the impulsive forces, F and -F, between the cannon and projectile are internal to the system and will therefore cancel from the analysis. Furthermore, during the time t = 0.03 s, the two recoil springs which are attached to the support each exert a nonimpulsive force Fs on the cannon. This is because t is very short, so that during this time the cannon only moves through a very small distance s. Consequently, Fs = ks  0, where k is the spring’s stiffness, which is also considered to be relatively small. Hence it can be concluded that momentum for the system is conserved in the horizontal direction.

15

Conservation of Linear Momentum. + ) (S m c(vc )1 + m p(vp)1 = -m c (vc )2 + m p(vp)2 0 + 0 = - (600 kg) (vc)2 + (4 kg )(vp)2 (vp)2 = 150 (vc)2

(1)

These unknown velocities are measured by a fixed observer. As in Example 15–7, they can also be related using the relative velocity equation. + (vp)2 = (vc)2 + vp>c S (vp)2 = -(vc)2 + 450 m>s

vc

vp x

2Fs

F

F (b)

Fig. 15–12

(2)

Solving Eqs. (1) and (2) yields (vc)2 = 2.98 m>s Ans. (vp)2 = 447.0 m>s Apply the principle of impulse and momentum to the projectile (or the cannon) and show that the average impulsive force on the projectile is 59.6 kN. NOTE: If the cannon is firmly fixed to its support (no springs), the

reactive force of the support on the cannon must be considered as an external impulse to the system, since the support would allow no movement of the cannon. In this case momentum is not conserved. *Only horizontal forces are shown on the free-body diagram.

15.3

Conservation of linear momentum for a system of partiCles

261

FuNDaMeNtaL prObLeMs F15–7. The freight cars A and B have a mass of 20 Mg and 15 Mg, respectively. Determine the velocity of A after collision if the cars collide and rebound, such that B moves to the right with a speed of 2 m>s. If A and B are in contact for 0.5 s, find the average impulsive force which acts between them.

A

3 m/s

B

F15–10. The spring is fixed to block A and block B is pressed against the spring. If the spring is compressed s = 200 mm and then the blocks are released, determine their velocity at the instant block B loses contact with the spring. The masses of blocks A and B are 10 kg and 15 kg, respectively. k  5 kN/m

1.5 m/s A

B

15

Prob. F15–7 Prob. F15–10 F15–8. The cart and package have a mass of 20 kg and 5 kg, respectively. If the cart has a smooth surface and it is initially at rest, while the velocity of the package is as shown, determine the final common velocity of the cart and package after the impact.

10 m/s 3

F15–11. Blocks A and B have a mass of 15 kg and 10 kg, respectively. If A is stationary and B has a velocity of 15 m>s just before collision, and the blocks couple together after impact, determine the maximum compression of the spring. 15 m/s

k  10 kN/m

5 4

A

B

Prob. F15–11 Prob. F15–8

F15–9. The 5-kg block A has an initial speed of 5 m>s as it slides down the smooth ramp, after which it collides with the stationary block B of mass 8 kg. If the two blocks couple together after collision, determine their common velocity immediately after collision. vA  5 m/s

F15–12. The cannon and support without a projectile have a mass of 250 kg. If a 20-kg projectile is fired from the cannon with a velocity of 400 m>s, measured relative to the cannon, determine the speed of the projectile as it leaves the barrel of the cannon. Neglect rolling resistance.

30

A 1.5 m B

Prob. F15–9

Prob. F15–12

262

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

prObLeMs 15–35. The 5-Mg bus B is traveling to the right at 20 m>s. Meanwhile a 2-Mg car A is traveling at 15 m>s to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

vB  20 m/s B

vA  15 m/s A

15

15–38. A railroad car having a mass of 15 Mg is coasting at 1.5 m>s on a horizontal track. At the same time another car having a mass of 12 Mg is coasting at 0.75 m>s in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred, and explain qualitatively what happened to this energy. 15–39. A ballistic pendulum consists of a 4-kg wooden block originally at rest, u = 0°. When a 2-g bullet strikes and becomes embedded in it, it is observed that the block swings upward to a maximum angle of u = 6°. Estimate the initial speed of the bullet.

Prob. 15–35

*15–36. The 50-kg boy jumps on the 5-kg skateboard with a horizontal velocity of 5 m>s. Determine the distance s the boy reaches up the inclined plane before momentarily coming to rest. Neglect the skateboard’s rolling resistance.

1.25 m

u

u

1.25 m

Prob. 15–39

s 30

Prob. 15–36

15–37. The 2.5-Mg pickup truck is towing the 1.5-Mg car using a cable as shown. If the car is initially at rest and the truck is coasting with a velocity of 30 km>h when the cable is slack, determine the common velocity of the truck and the car just after the cable becomes taut. Also, find the loss of energy.

*15–40. The boy B jumps off the canoe at A with a velocity 5 m>s relative to the canoe as shown. If he lands in the second canoe C, determine the final speed of both canoes after the motion. Each canoe has a mass of 40 kg. The boy’s mass is 30 kg, and the girl D has a mass of 25 kg. Both canoes are originally at rest.

vB/A B

5 m/s

30 D

30 km/h

Prob. 15–37

A

C

Prob. 15–40

15.3

263

Conservation of linear momentum for a system of partiCles

15–41. The block of mass m travels at v1 in the direction u1 shown at the top of the smooth slope. Determine its speed v2 and its direction u2 when it reaches the bottom.

15–43. The 20-g bullet is traveling at 400 m>s when it becomes embedded in the 2-kg stationary block. Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the block and the plane is mk = 0.2.

z

400 m/s v1 y

x

u1

Prob. 15–43

h

15

v2 u2

Prob. 15–41

15–42. A toboggan having a mass of 10 kg starts from rest at A and carries a girl and boy having a mass of 40 kg and 45 kg, respectively. When the toboggan reaches the bottom of the slope at B, the boy is pushed off from the back with a horizontal velocity of vb>t = 2 m>s, measured relative to the toboggan. Determine the velocity of the toboggan afterwards. Neglect friction in the calculation.

*15–44. A 4-kg projectile travels with a horizontal velocity of 600 m>s before it explodes and breaks into two fragments A and B of mass 1.5 kg and 2.5 kg, respectively. If the fragments travel along the parabolic trajectories shown, determine the magnitude of velocity of each fragment just after the explosion and the horizontal distance dA where segment A strikes the ground at C.

15–45. A 4-kg projectile travels with a horizontal velocity of 600 m>s before it explodes and breaks into two fragments A and B of mass 1.5 kg and 2.5 kg, respectively. If the fragments travel along the parabolic trajectories shown, determine the magnitude of velocity of each fragment just after the explosion and the horizontal distance dB where segment B strikes the ground at D.

vA

A 600 m/s

45

3m B

vb/t

vt

B

30 vB

60 m D

C dA

Prob. 15–42

A

dB

Probs. 15–44/45

264

Chapter 15

KinetiCs

of a

partiCle: impulse

15–46. The 10-Mg barge B supports a 2-Mg automobile A. If someone drives the automobile to the other side of the barge, determine how far the barge moves. Neglect the resistance of the water.

and

momentum

15–49. The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the 10-g bullet is traveling at 300 m>s when it becomes embedded in the 10-kg block, determine the distance the block will slide up along the plane before momentarily stopping.

300 m/s 40 m

A A

B

15

30

Prob. 15–49

Prob. 15–46

15–47. Block A has a mass of 2 kg and slides into an open ended box B with a velocity of 2 m>s. If the box B has a mass of 3 kg and rests on top of a plate P that has a mass of 3 kg, determine the distance the plate moves after it stops sliding on the floor. Also, how long is it after impact before all motion ceases? The coefficient of kinetic friction between the box and the plate is mk = 0.2, and between the plate and the floor mk = 0.4. Also, the coefficient of static friction between the plate and the floor is ms = 0.5.

*15–48. Block A has a mass of 2 kg and slides into an open ended box B with a velocity of 2 m>s. If the box B has a mass of 3 kg and rests on top of a plate P that has a mass of 3 kg, determine the distance the plate moves after it stops sliding on the floor. Also, how long is it after impact before all motion ceases? The coefficient of kinetic friction between the box and the plate is mk = 0.2, and between the plate and the floor mk = 0.1. Also, the coefficient of static friction between the plate and the floor is ms = 0.12.

2 m/s

Probs. 15–47/48

A

1.25 m

vc

B

vb/c

Prob. 15–50

15–51. The 30-Mg freight car A and 15-Mg freight car B are moving towards each other with the velocities shown. Determine the maximum compression of the spring mounted on car A. Neglect rolling resistance.

20 km/h

B

A

15–50. The cart has a mass of 3 kg and rolls freely from A down the slope. When it reaches the bottom, a spring loaded gun fires a 0.5-kg ball out the back with a horizontal velocity of vb>c = 0.6 m>s, measured relative to the cart. Determine the final velocity of the cart.

A

10 km/h k  3 MN/m

P

Prob. 15–51

B

15.3

Conservation of Linear MoMentuM for a systeM of PartiCLes

*15–52. The two blocks A and B each have a mass of 5 kg and are suspended from parallel cords. A spring, having a stiffness of k = 60 N>m, is attached to B and is compressed 0.3 m against A and B as shown. Determine the maximum angles u and f of the cords when the blocks are released from rest and the spring becomes unstretched.

2m

u

2m

A

265

15–55. Block A has a mass of 5 kg and is placed on the smooth triangular block B having a mass of 30 kg. If the system is released from rest, determine the distance B moves from point O when A reaches the bottom. Neglect the size of block A. *15–56. Solve Prob. 15–55 if the coefficient of kinetic friction between A and B is μk = 0.3. Neglect friction between block B and the horizontal plane.

f A

B

Prob. 15–52 B 30

O

15–53. Blocks A and B have masses of 40 kg and 60 kg, respectively. They are placed on a smooth surface and the spring connected between them is stretched 2 m. If they are released from rest, determine the speeds of both blocks the instant the spring becomes unstretched.

k  180 N/m A

0.5 m

Probs. 15–55/56 15–57. The free-rolling ramp has a mass of 40 kg. A 10-kg crate is released from rest at A and slides down 3.5 m to point B. If the surface of the ramp is smooth, determine the ramp’s speed when the crate reaches B. Also, what is the velocity of the crate?

B

Prob. 15–53

15–54. Two boxes A and B, each having a mass of 80 kg, sit on the 250-kg conveyor which is free to roll on the ground. If the belt starts from rest and begins to run with a speed of 1 m>s, determine the final speed of the conveyor if (a) the boxes are not stacked and A falls off then B falls off, and (b) A is stacked on top of B and both fall off together.

B

A

Prob. 15–54

3.5 m

A

30

Prob. 15–57

B

15

266

Chapter 15

KinetiCs

of a

15.4

Plane of contact A

vA

B

vB

Line of impact

Central impact (a) Plane of contact A

15

B

u

Line of impact f

partiCle: impulse

and

momentum

Impact

Impact occurs when two bodies collide with each other during a very short period of time, causing relatively large (impulsive) forces to be exerted between the bodies. The striking of a hammer on a nail, or a golf club on a ball, are common examples of impact loadings. In general, there are two types of impact. Central impact occurs when the direction of motion of the mass centers of the two colliding particles is along a line passing through the mass centers of the particles. This line is called the line of impact, which is perpendicular to the plane of contact, Fig. 15–13a. When the motion of one or both of the particles make an angle with the line of impact, Fig. 15–13b, the impact is said to be oblique impact.

Central Impact. To illustrate the method for analyzing the mechanics

Oblique impact

of impact, consider the case involving the central impact of the two particles A and B shown in Fig. 15–14.

(b)

• The particles have the initial momenta shown in Fig. 15–14a.

vA

vB

Fig. 15–13



• • mA(vA)1

mB(vB)1

Require A (v )  (v ) B A 1 B 1 Before impact



(a)

P dt A Effect of A on B

P dt

Provided (vA)1 7 (vB)1 , collision will eventually occur. During the collision the particles must be thought of as deformable or nonrigid. The particles will undergo a period of deformation such that they exert an equal but opposite deformation impulse 1 P dt on each other, Fig. 15–14b. Only at the instant of maximum deformation will both particles move with a common velocity v, since their relative motion is zero, Fig. 15–14c. Afterward a period of restitution occurs, in which case the particles will either return to their original shape or remain permanently deformed. The equal but opposite restitution impulse 1 R dt pushes the particles apart from one another, Fig. 15–14d. In reality, the physical properties of any two bodies are such that the deformation impulse will always be greater than that of restitution, i.e., 1 P dt 7 1 R dt. Just after separation the particles will have the final momenta shown in Fig. 15–14e, where (vB)2 7 (vA)2 .

v

R dt

A Effect of A on B

B Effect of B on A

Deformation impulse

R dt

A B

mB(vB)2

B Effect of B on A

Restitution impulse

Maximum deformation (b)

mA(vA)2

A (vB)2  (vA)2 B After impact

(d)

(c)

Fig. 15–14

(e)

15.4

impaCt

267

In most problems the initial velocities of the particles will be known, and it will be necessary to determine their final velocities (vA)2 and (vB)2 . In this regard, momentum for the system of particles is conserved since during collision the internal impulses of deformation and restitution cancel. Hence, referring to Fig. 15–14a and Fig. 15–14e we require + ) (S

m A (vA)1 + m B(vB)1 = m A(vA)2 + m B(vB)2

(15–10)

In order to obtain a second equation necessary to solve for (vA)2 and (vB)2 , we must apply the principle of impulse and momentum to each particle. For example, during the deformation phase for particle A, Figs. 15–14a, 15–14b, and 15–14c, we have + ) (S

m A (vA)1 -

L

15

P dt = m Av

For the restitution phase, Figs. 15–14c, 15–14d, and 15–14e, + ) (S

mAv -

L

R dt = m A (vA)2

The ratio of the restitution impulse to the deformation impulse is called the coefficient of restitution, e. From the above equations, this value for particle A is

e =

L

L

R dt = P dt

v - (vA)2 (vA)1 - v

In a similar manner, we can establish e by considering particle B, Fig. 15–14. This yields

e =

L

L

R dt = P dt

(vB)2 - v v - (vB)1

If the unknown v is eliminated from the above two equations, the coefficient of restitution can be expressed in terms of the particles’ initial and final velocities as

+ ) (S

e =

(vB)2 - (vA)2 (vA)1 - (vB)1

(15–11)

268

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

Provided a value for e is specified, Eqs. 15–10 and 15–11 can be solved simultaneously to obtain (vA)2 and (vB)2 . In doing so, however, it is important to carefully establish a sign convention for defining the positive direction for both vA and vB and then use it consistently when writing both equations. As noted from the application shown, and indicated symbolically by the arrow in parentheses, we have defined the positive direction to the right when referring to the motions of both A and B. Consequently, if a negative value results from the solution of either (vA)2 or (vB)2 , it indicates motion is to the left. The quality of a manufactured tennis ball is measured by the height of its bounce, which can be related to its coefficient of restitution. Using the mechanics of oblique impact, engineers can design a separation device to 15 remove substandard tennis balls from a production line. (© Gary S. Settles/Science Source)

Coefficient of Restitution.

From Figs. 15–14a and 15–14e, it is seen that Eq. 15–11 states that e is equal to the ratio of the relative velocity of the particles’ separation just after impact, (vB) 2 - (vA) 2 , to the relative velocity of the particles’ approach just before impact, (vA)1 - (vB)1 . By measuring these relative velocities experimentally, it has been found that e varies appreciably with impact velocity as well as with the size and shape of the colliding bodies. For these reasons the coefficient of restitution is reliable only when used with data which closely approximate the conditions which were known to exist when measurements of it were made. In general e has a value between zero and one, and one should be aware of the physical meaning of these two limits.

Elastic Impact (e  1). If the collision between the two particles is perfectly elastic, the deformation impulse 11 P dt 2 is equal and opposite to the restitution impulse 11 R dt 2 . Although in reality this can never be achieved, e = 1 for an elastic collision. when e = 0. In this case there is no restitution impulse 11 R dt = 0 2 , so that after collision both particles couple or stick together and move with a common velocity. From the above derivation it should be evident that the principle of work and energy cannot be used for the analysis of impact problems since it is not possible to know how the internal forces of deformation and restitution vary or displace during the collision. By knowing the particle’s velocities before and after collision, however, the energy loss during collision can be calculated on the basis of the difference in the particle’s kinetic energy. This energy loss, U1 - 2 = T 2 - T 1 , occurs because some of the initial kinetic energy of the particle is transformed into thermal energy as well as creating sound and localized deformation of the material when the collision occurs. In particular, if the impact is perfectly elastic, no energy is lost in the collision; whereas if the collision is plastic, the energy lost during collision is a maximum.

Plastic Impact (e  0). The impact is said to be inelastic or plastic

The mechanics of pool depends upon application of the conservation of momentum and the coefficient of restitution.

15.4

269

impaCt

y

procedure for analysis (Central impact)

(vA)2

In most cases the final velocities of two smooth particles are to be determined just after they are subjected to direct central impact. Provided the coefficient of restitution, the mass of each particle, and each particle’s initial velocity just before impact are known, the solution to this problem can be obtained using the following two equations:

A

u2

(vB)2 B

u1

f2

Line of impact

x

f1

(vB)1 (vA)1 Plane of contact (a)

• The conservation of momentum applies to the system of particles, mv 1 = mv 2 .

• The coefficient of restitution, e = [(vB ) 2 - (vA ) 2]>[(vA ) 1 - (vB ) 1],

relates the relative velocities of the particles along the line of impact, just before and just after collision. When applying these two equations, the sense of an unknown velocity can be assumed. If the solution yields a negative magnitude, the velocity acts in the opposite sense.

Oblique Impact. When oblique impact occurs between two smooth particles, the particles move away from each other with velocities having unknown directions as well as unknown magnitudes. Provided the initial velocities are known, then four unknowns are present in the problem. As shown in Fig. 15–15a, these unknowns may be represented either as (vA) 2 , (vB) 2 , u 2 , and f2 , or as the x and y components of the final velocities.

mA(vAy)2 mA(vAx)1

A

A

Fdt

mA(vAx)2

A

mA(vAy)1 mB(vBy)2 B

mB(vBx)1

Fdt

B

B

mB(vBx)2

mB(vBy)1 (b)

Fig. 15–15

procedure for analysis (Oblique impact) If the y axis is established within the plane of contact and the x axis along the line of impact, the impulsive forces of deformation and restitution act only in the x direction, Fig. 15–15b. By resolving the velocity or momentum vectors into components along the x and y axes, Fig. 15–15b, it is then possible to write four independent scalar equations in order to determine (vAx) 2 , (vAy) 2 , (vBx) 2 , and (vBy) 2 .

• Momentum of the system is conserved along the line of impact, x axis, so that m(vx) 1 = m(vx) 2 . • The coefficient of restitution, e = [(vBx) 2 - (vAx) 2]>[(vAx) 1 - (vBx) 1], relates the relative-velocity • • •

components of the particles along the line of impact (x axis). If these two equations are solved simultaneously, we obtain (vAx) 2 and (vBx) 2. Momentum of particle A is conserved along the y axis, perpendicular to the line of impact, since no impulse acts on particle A in this direction. As a result m A (vAy) 1 = mA(vAy) 2 or (vAy) 1 = (vAy) 2 Momentum of particle B is conserved along the y axis, perpendicular to the line of impact, since no impulse acts on particle B in this direction. Consequently (vBy) 1 = (vBy) 2.

Application of these four equations is illustrated in Example 15.11.

15

270

Chapter 15

KinetiCs

of a

and

momentum

15.9

ExaMPlE

A

u

1m B Line of impact (a)

15

Datum

0

3(9.81)N 1 m

1

The bag A, having a mass of 3 kg, is released from rest at the position u = 0, as shown in Fig. 15–16a. After falling to u = 90, it strikes an 9-kg box B. If the coefficient of restitution between the bag and box is e = 0.5, determine the velocities of the bag and box just after impact. What is the loss of energy during collision? SolutIon This problem involves central impact. Why? Before analyzing the mechanics of the impact, however, it is first necessary to obtain the velocity of the bag just before it strikes the box. Conservation of energy. With the datum at u = 0, Fig. 15–16b, we have T0 + V 0 = T1 + V 1 1 0 + 0 = (3 kg)(vA)21 - 3 kg(9.81 m>s2)(1 m); (vA)1 = 4.429 m>s 2 Conservation of Momentum. After impact we will assume A and B travel to the left. Applying the conservation of momentum to the system, Fig. 15–16c, we have + ) (d m B (vB)1 + m A(vA)1 = m B(vB)2 + m A(vA)2 0 + (3 kg)(4.429 m>s) = (9 kg)(yB)2 = (3 kg)(yA)2 (vA)2 = 4.429 - 3(vB)2

3(9.81)N (b)

(vB)1  0

partiCle: impulse

Coefficient of restitution. Realizing that for separation to occur after collision (vB)2 7 (vA)2 , Fig. 15–16c, we have (vB)2 - (vA)2 (vB)2 - (vA)2 + ) (d e = ; 0.5 = (vA)1 - (vB)1 4.429 m>s - 0 (vA)2 = (vB)2 - 2.2145 (2)

A

B

Solving Eqs. 1 and 2 simultaneously yields

(vA)1  4.429 m/s Just before impact

A

B

(vB)2 Just after impact (c)

Fig. 15–16

(1)

(vA)2

(vA)2 = - 0.554 m>s = 0.554 m>s S

and (vB)2 = 1.66 m>s d Ans.

Loss of energy. Applying the principle of work and energy to the bag and box just before and just after collision, we have U1 - 2 = T 2 - T 1; 1 1 U1 - 2 = c (9 kg)(1.661 m>s)2 + (3 kg)(0.554 m>s)2 d 2 2 1 - c (3 kg)(4.429 m>s)2 d 2 U1 - 2 = -16.6 J Ans. NOTE: The energy loss occurs due to inelastic deformation during the

collision.

15.4

ExamplE

271

Impact

15.10

Ball B shown in Fig. 15–17a has a mass of 1.5 kg and is suspended from the ceiling by a 1-m-long elastic cord. If the cord is stretched downward 0.25 m and the ball is released from rest, determine how far the cord stretches after the ball rebounds from the ceiling. The stiffness of the cord is k = 800 N>m, and the coefficient of restitution between the ball and ceiling is e = 0.8. The ball makes a central impact with the ceiling.

1

Datum k  800 N/m

y  (1  0.25) m

B 0

Solution First we must obtain the velocity of the ball just before it strikes the ceiling using energy methods, then consider the impulse and momentum between the ball and ceiling, and finally again use energy methods to determine the stretch in the cord. Conservation of Energy. With the datum located as shown in Fig. 15–17a, realizing that initially y = y 0 = (1 + 0.25) m = 1.25 m, we have T0 + V 0 = T1 + V 1 1 1 1 2 2 2 2 m(vB)0 - WB y0 + 2 ks = 2 m(vB)1 + 0 1 2 0 - 1.5(9.81)N(1.25 m) + 2(800 N>m)(0.25 m) = 12(1.5 kg)(vB)21

(a)

15

y x

(vB)1 = 2.968 m>s c The interaction of the ball with the ceiling will now be considered using the principles of impact.* Since an unknown portion of the mass of the ceiling is involved in the impact, the conservation of momentum for the ball–ceiling system will not be written. The “velocity” of this portion of ceiling is zero since it (or the earth) are assumed to remain at rest both before and after impact. Coefficient of Restitution. Fig. 15–17b. (vB)2 - 0 (vB)2 - (vA)2 (+ c ) e = ; 0.8 = (vA)1 - (vB)1 0 - 2.968 m>s (vB)2 = -2.374 m>s = 2.374 m>s T Conservation of Energy. The maximum stretch s3 in the cord can be determined by again applying the conservation of energy equation to the ball just after collision. Assuming that y = y 3 = (1 + s3) m, Fig. 15–17c, then T2 + V 2 = T3 + V 3 1 1 1 2 2 2 2 m(vB)2 + 0 = 2 m(vB)3 - WBy3 + 2 ks3 1 1 2 2 2 (1.5 kg)(2.37 m>s) = 0 - 9.81(1.5)N(1 m + s3) + 2 (800 N>m)s3 400s23 - 14.715s3 - 18.94 = 0 Solving this quadratic equation for the positive root yields s3 = 0.237 m = 237 mm Ans. *The weight of the ball is considered a nonimpulsive force.

(vB)2

(vB)1  2.97 m/s (b)

Datum

2 B k  800 N/m y  (1  s3) m

(c)

Fig. 15–17

3

272

Chapter 15

KinetiCs

of a

y

(vB)1  1 m/s f1  45

A u1  30

x Line of impact

B

(vA)1  3 m/s

Plane of contact (a)

15

A

A

Fdt

A

mA(vAx)2

mA(vAy)2

mA(vAy)1 mB(vBy)1 mB(vBx)1

and

momentum

15.11

ExamplE

mA(vAx)1

partiCle: impulse

Fdt

B

B

Two smooth disks A and B, having a mass of 1 kg and 2 kg, respectively, collide with the velocities shown in Fig. 15–18a. If the coefficient of restitution for the disks is e = 0.75, determine the x and y components of the final velocity of each disk just after collision. Solution This problem involves oblique impact. Why? In order to solve it, we have established the x and y axes along the line of impact and the plane of contact, respectively, Fig. 15–18a. Resolving each of the initial velocities into x and y components, we have (vA x )1 = 3 cos 30 = 2.598 m>s (vA y )1 = 3 sin 30 = 1.50 m>s (vBx)1 = -1 cos 45 = -0.7071 m>s (vBy)1 = -1 sin 45 = -0.7071 m>s The four unknown velocity components after collision are assumed to act in the positive directions, Fig. 15–18b. Since the impact occurs in the x direction (line of impact), the conservation of momentum for both disks can be applied in this direction. Why? Conservation of “x” Momentum. In reference to the momentum diagrams, we have + ) (S m A (vA x )1 + m B (vBx )1 = m A (vA x )2 + m B (vBx )2 1 kg(2.598 m>s) + 2 kg(-0.707 m>s) = 1 kg(vA x )2 + 2 kg(vBx )2

mB(vBx)2

B

(vA x )2 + 2(vBx )2 = 1.184

(1)

Coefficient of Restitution (x). mB(vBy)2

+ ) (S

(b)

e =

(vBx )2 - (vA x )2 (vBx )2 - (vA x )2 ; 0.75 = 2.598 m>s - (-0.7071 m>s) (vA x )1 - (vBx )1 (vBx )2 - (vA x )2 = 2.482

Solving Eqs. 1 and 2 for (vA x )2 and (vBx )2 yields (vA x )2 = -1.26 m>s = 1.26 m>s d (vBx )2 = 1.22 m>s S (vA)2  1.96 m/s u2  50.0

A

(2) Ans.

Conservation of “y” Momentum. The momentum of each disk is conserved in the y direction (plane of contact), since the disks are smooth and therefore no external impulse acts in this direction. From Fig. 15–18b,

y

B f2  30.1

x

(vB)2  1.41 m/s (c)

Fig. 15–18

(+ c ) m A (vA y )1 = m A (vA y )2;

(vA y )2 = 1.50 m>s c

(+ c ) m B (vBy)1 = m B (vBy )2;

(vBy )2 = -0.707 m>s = 0.707 m>s T Ans.

Ans.

NOTE: Show that when the velocity components are summed vectorially, one obtains the results shown in Fig. 15–18c.

15.4

273

impaCt

FuNDaMeNtaL prObLeMs F15–13. Determine the coefficient of restitution e between ball A and ball B. The velocities of A and B before and after the collision are shown. 8 m/s

F15–16. The ball strikes the smooth wall with a velocity of (vb)1 = 20 m>s. If the coefficient of restitution between the ball and the wall is e = 0.75, determine the velocity of the ball just after the impact.

2 m/s A

(vb)2

B

u 30 Before collision 1 m/s

(vb)1  20 m/s

9 m/s A

B

After collision

Prob. F15–13

Prob. F15–16 15 F15–17. Disk A has a mass of 2 kg and slides on the smooth horizontal plane with a velocity of 3 m>s. Disk B has a mass of 11 kg and is initially at rest. If after impact A has a velocity of 1 m>s, parallel to the positive x axis, determine the speed of disk B after impact. y

F15–14. The 15-Mg tank car A and 25-Mg freight car B travel toward each other with the velocities shown. If the coefficient of restitution between the bumpers is e = 0.6, determine the velocity of each car just after the collision. x A

5 m/s

B

7 m/s

B 3 m/s A

Prob. F15–14 F15–15. The 15-kg package A has a speed of 1.5 m>s when it enters the smooth ramp. As it slides down the ramp, it strikes the 40-kg package B which is initially at rest. If the coefficient of restitution between A and B is e = 0.6, determine the velocity of B just after the impact. 1.5 m/s

Prob. F15–17 F15–18. Two disks A and B each have a mass of 1 kg and the initial velocities shown just before they collide. If the coefficient of restitution is e = 0.5, determine their speeds just after impact. y

A

(vA)1  4 m/s

3m

B

A x

1.5 m

(vB)1  3 m/s

B

3

5 4

Prob. F15–15

Prob. F15–18

274

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and

momentum

prObLeMs 15–58. Disk A has a mass of 250 g and is sliding on a smooth horizontal surface with an initial velocity (vA )1 = 2 m>s. It makes a direct collision with disk B, which has a mass of 175 g and is originally at rest. If both disks are of the same size and the collision is perfectly elastic (e = 1), determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same. 15–59. The 5-Mg truck and 2-Mg car are traveling with the free-rolling velocities shown just before they collide. After the collision, the car moves with a velocity of 15 km>h to the 15 right relative to the truck. Determine the coefficient of restitution between the truck and car and the loss of energy due to the collision.

15–61. Ball A has a mass of 3 kg and is moving with a velocity of 8 m>s when it makes a direct collision with ball B, which has a mass of 2 kg and is moving with a velocity of 4 m>s. If e = 0.7, determine the velocity of each ball just after the collision. Neglect the size of the balls. 8 m/s

4 m/s

A

B

Prob. 15–61 15–62. The 15-kg block A slides on the surface for which mk = 0.3. The block has a velocity v = 10 m>s when it is s = 4  m  from the 10-kg block B. If the unstretched spring has  a  stiffness k = 1000 N>m, determine the maximum compression of the spring due to the collision. Take e = 0.6.

30 km/h 10 km/h

10 m/s

k  1000 N/m

A

B

Prob. 15–59 s

Prob. 15–62 *15–60. Disk A has a mass of 2 kg and is sliding forward on the smooth surface with a velocity (vA )1 = 5 m>s when it strikes the 4-kg disk B, which is sliding towards A at (vB )1 = 2 m>s, with direct central impact. If the coefficient of restitution between the disks is e = 0.4, compute the velocities of A and B just after collision.

15–63. The four smooth balls each have the same mass m. If A and B are rolling forward with velocity v and strike C, explain why after collision C and D each move off with velocity v. Why doesn’t D move off with velocity 2v? The collision is elastic, e = 1. Neglect the size of each ball. *15–64. The four balls each have the same mass m. If A and  B are rolling forward with velocity v and strike C, determine the velocity of each ball after the first three collisions. Take e = 0.5 between each ball.

(vA)1  5 m/s

(vB)1  2 m/s

A

B

Prob. 15–60

v

v

A

B

Probs. 15–63/64

C D

15.4 15–65. Two smooth spheres A and B each have a mass m. If A is given a velocity of v0, while sphere B is at rest, determine the velocity of B just after it strikes the wall. The coefficient of restitution for any collision is e.

impaCt

275

*15–68. A 1-kg ball A is traveling horizontally at 20 m>s when it strikes a 10-kg block B that is at rest. If the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic friction between the plane and the block is mk = 0.4, determine the time for the block B to stop sliding. 15–69. The three balls each have a mass m. If A has a speed v just before a direct collision with B, determine the speed of C after collision. The coefficient of restitution between each pair of balls is e. Neglect the size of each ball.

v0

A

B

Prob. 15–65 15–66. A pitching machine throws the 0.5-kg ball toward the wall with an initial velocity vA = 10 m>s as shown. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall if e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C.

15

v B

C

A

Prob. 15–69

B

15–70. Block A, having a mass m, is released from rest, falls a distance h and strikes the plate B having a mass 2m. If the coefficient of restitution between A and B is e, determine the velocity of the plate just after collision. The spring has a stiffness k.

vA  10 m/s

A

30 1.5 m

C 3m s

Prob. 15–66 15–67. A 300-g ball is kicked with a velocity of vA = 25 m>s at point A as shown. If the coefficient of restitution between the ball and the field is e = 0.4, determine the magnitude and direction u of the velocity of the rebounding ball at B.

A h B

vA  25 m/s

v¿B

k

30

u B

A

Prob. 15–67

Prob. 15–70

276

Chapter 15

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15–71. The cue ball A is given an initial velocity (vA)1 = 5 m>s. If it makes a direct collision with ball B (e = 0.8), determine the velocity of B and the angle u just after it rebounds from the cushion at C (e = 0.6). Each ball has a mass of 0.4 kg. Neglect their size.

(vA)1  5 m/s

B

momentum

15–74. The pile P has a mass of 800 kg and is being driven into loose sand using the 300-kg hammer C which is dropped a distance of 0.5 m from the top of the pile. Determine the initial speed of the pile just after it is struck by the hammer. The coefficient of restitution between the hammer and the pile is e = 0.1. Neglect the impulses due to the weights of the pile and hammer and the impulse due to the sand during the impact. 15–75. The pile P has a mass of 800 kg and is being driven into loose sand using the 300-kg hammer C which is dropped a distance of 0.5 m from the top of the pile. Determine the distance the pile is driven into the sand after one blow if the sand offers a frictional resistance against the pile of 18 kN. The coefficient of restitution between the hammer and the pile is e = 0.1. Neglect the impulses due to the weights of the pile and hammer and the impulse due to the sand during the impact.

A 30

C u

15

and

Prob. 15–71 *15–72. The two disks A and B have a mass of 3 kg and 5  kg, respectively. If they collide with the initial velocities shown, determine their velocities just after impact. The coefficient of restitution is e = 0.65.

C 0.5 m P

(vB)1  7 m/s B

A

(vA)1  6 m/s

60

Line of impact

Prob. 15–72 15–73. The 0.5-kg ball is fired from the tube at A with a velocity of v = 6 m>s. If the coefficient of restitution between the ball and the surface is e = 0.8, determine the height h after it bounces off the surface. v  6 m/s

Probs. 15–74/75

*15–76. A ball of mass m is dropped vertically from a height h0 above the ground. If it rebounds to a height of h1, determine the coefficient of restitution between the ball and the ground.

30

A C

2m

h0 h1

h B

Prob. 15–73

Prob. 15–76

15.4 15–77. Two smooth disks A and B each have a mass of 0.5  kg. If both disks are moving with the velocities shown when they collide, determine their final velocities just after collision. The coefficient of restitution is e = 0.75. 15–78. Two smooth disks A and B each have a mass of 0.5 kg. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision B travels along a line, 30° counterclockwise from the y axis.

277

impaCt

*15–80. The 2-kg ball is thrown at the suspended 20-kg block with a velocity of 4 m>s. If the coefficient of restitution between the ball and the block is e = 0.8, determine the maximum height h to which the block will swing before it momentarily stops. 15–81. The 2-kg ball is thrown at the suspended 20-kg block with a velocity of 4 m>s. If the time of impact between the ball and the block is 0.005 s, determine the average normal force exerted on the block during this time. Take e = 0.8.

y

15 (vA)1  6 m/s

4 m/s

x B

B

A

5 4 3

h

A

Probs. 15–80/81 15–82. The three balls each have the same mass m. If A is released from rest at u, determine the angle f to which C rises after collision. The coefficient of restitution between each ball is e.

(vB)1  4 m/s

Probs. 15–77/78

l

15–79. A ball of negligible size and mass m is given a velocity of v0 on the center of the cart which has a mass M and is originally at rest. If the coefficient of restitution between the ball and walls A and B is e, determine the velocity of the ball and the cart just after the ball strikes A. Also, determine the total time needed for the ball to strike A, rebound, then strike B, and rebound and then return to the center of the cart. Neglect friction.

l

u

f

l

A B C

Prob. 15–82 15–83. The girl throws the 0.5-kg ball toward the wall with an initial velocity vA = 10 m>s. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall if the coefficient of restitution e = 0.5, and (c) the distance s from the wall to where it strikes the ground at C. B

v0

vA  10 m/s

A

A

30 1.5 m

B C

d

d

Prob. 15–79

s

3m

Prob. 15–83

278

Chapter 15

KinetiCs

partiCle: impulse

of a

*15–84. The 1-kg ball is dropped from rest at point A, 2 m above the smooth plane. If the coefficient of restitution between the ball and the plane is e = 0.6, determine the distance d where the ball again strikes the plane.

and

momentum

15–86. Two smooth billiard balls A and B each have a mass of 200 g. If A strikes B with a velocity (vA)1 = 1.5 m>s as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.85. Neglect the size of each ball.

A

y A

2m B

(vA)1  1.5 m/s

15 40

30

x

B d

C

Prob. 15–86 Prob. 15–84

15–85. Disks A and B have a mass of 15 kg and 10 kg, respectively. If they are sliding on a smooth horizontal plane with the velocities shown, determine their speeds just after impact. The coefficient of restitution between them is e = 0.8.

y 5

4

15–87. A ball is thrown onto a rough floor at an angle u. If it rebounds at an angle f and the coefficient of kinetic friction is m, determine the coefficient of restitution e. Neglect the size of the ball. Hint: Show that during impact, the average impulses in the x and y directions are related by Ix = mIy . Since the time of impact is the same, Fx t = mFy t or Fx = mFy . *15–88. A ball is thrown onto a rough floor at an angle of u = 45. If it rebounds at the same angle f = 45, determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is e = 0.6. Hint: Show that during impact, the average impulses in the x and y directions are related by Ix = mIy . Since the time of impact is the same, Fx t = mFy t or Fx = mFy .

Line of impact

3

A

10 m/s y

x 8 m/s

B

f

u

x

Prob. 15–85

Probs. 15–87/88

15.4 15–89. The two billiard balls A and B are originally in contact with one another when a third ball C strikes each of them at the same time as shown. If ball C remains at rest after the collision, determine the coefficient of restitution. All the balls have the same mass. Neglect the size of each ball.

279

*15–92. Two smooth disks A and B have the initial velocities shown just before they collide. If they have masses m A = 4 kg and m B = 2 kg, determine their speeds just after impact. The coefficient of restitution is e = 0.8.

A

5

4

3

v

A C B

15

vB  8 m/s

15–90. Disks A and B have masses of 2 kg and 4 kg, respectively. If they have the velocities shown, and e = 0.4, determine their velocities just after direct central impact.

2 m/s

vA  15 m/s

B

Prob. 15–89

(vA)1

impaCt

(vB)1

A

5 m/s

B

Prob. 15–92

15–93. The 200-g billiard ball is moving with a speed of 2.5 m>s when it strikes the side of the pool table at A. If the coefficient of restitution between the ball and the side of the table is e = 0.6, determine the speed of the ball just after striking the table twice, i.e., at A, then at B. Neglect the size of the ball.

Prob. 15–90 15–91. If disk A is sliding along the tangent to disk B and strikes B with a velocity v, determine the velocity of B after  the collision and compute the loss of kinetic energy during the collision. Neglect friction. Disk B is originally at rest. The coefficient of restitution is e, and each disk has the same size and mass m.

B

v  2.5 m/s

45

v

A

Prob. 15–91

Prob. 15–93

A B

280

Chapter 15

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of a

partiCle: impulse

15.5

and

momentum

Angular Momentum

The angular momentum of a particle about point O is defined as the “moment” of the particle’s linear momentum about O. Since this concept is analogous to finding the moment of a force about a point, the angular momentum, HO , is sometimes referred to as the moment of momentum.

z

HO

Scalar Formulation.

O

y

d mv

15

If a particle moves along a curve lying in the x–y plane, Fig. 15–19, the angular momentum at any instant can be determined about point O (actually the z axis) by using a scalar formulation. The magnitude of HO is

(15–12)

(HO)z = (d)(mv)

x

Fig. 15–19

Here d is the moment arm or perpendicular distance from O to the line of action of mv. Common units for (HO)z are kg # m2 >s. The direction of HO is defined by the right-hand rule. As shown, the curl of the fingers of the right hand indicates the sense of rotation of mv about O, so that in this case the thumb (or HO) is directed perpendicular to the x–y plane along the +z axis. z

Vector Formulation.

If the particle moves along a space curve, Fig. 15–20, the vector cross product can be used to determine the angular momentum about O. In this case

mv r O

y

HO  r  mv x

Fig. 15–20

HO = r * mv

(15–13)

Here r denotes a position vector drawn from point O to the particle. As shown in the figure, HO is perpendicular to the shaded plane containing r and mv. In order to evaluate the cross product, r and mv should be expressed in terms of their Cartesian components, so that the angular momentum can be determined by evaluating the determinant:

i HO = 3 rx mvx

j ry mvy

k rz 3 mvz

(15–14)

15.6

15.6

281

relation Between moment of a forCe and angular momentum

Relation Between Moment of a Force and Angular Momentum

The moments about point O of all the forces acting on the particle in Fig. 15–21a can be related to the particle’s angular momentum by applying the equation of motion. If the mass of the particle is constant, we may write

z

# F = mv

F

r

The moments of the forces about point O can be obtained by performing a cross-product multiplication of each side of this equation by the position vector r, which is measured from the x, y, z inertial frame of reference. We have

y O

x

Inertial coordinate system (a)

# MO = r * F = r * mv

Fig. 15–21

From Appendix B, the derivative of r * mv can be written as # d # # HO = (r * mv) = r * mv + r * mv dt # # # The first term on the right side, r * mv = m(r * r) = 0, since the cross product of a vector with itself is zero. Hence, the above equation becomes # MO = HO

(15–15)

which states that the resultant moment about point O of all the forces acting on the particle is equal to the time rate of change of the particle’s angular momentum about point O. This result is similar to Eq. 15–1, i.e., # F = L

(15–16)

Here L = mv, so that the resultant force acting on the particle is equal to the time rate of change of the particle’s linear momentum. From the derivations, it is seen that Eqs. 15–15 and 15–16 are actually another way of stating Newton’s second law of motion. In other sections of this book it will be shown that these equations have many practical applications when extended and applied to problems involving either a system of particles or a rigid body.

15

282

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

System of Particles.

An equation having the same form as Eq. 15–15 may be derived for the system of particles shown in Fig. 15–21b. The forces acting on the arbitrary ith particle of the system consist of a resultant external force Fi and a resultant internal force fi . Expressing the moments of these forces about point O, using the form of Eq. 15–15, we have # (ri * Fi) + (ri * fi) = (Hi)O

15

# Here (Hi)O is the time rate of change in the angular momentum of the ith particle about O. Similar equations can be written for each of the other particles of the system. When the results are summed vectorially, the result is # (ri * Fi) + (ri * fi) = (Hi)O The second term is zero since the internal forces occur in equal but opposite collinear pairs, and hence the moment of each pair about point O is zero. Dropping the index notation, the above equation can be written in a simplified form as # MO = HO

(15–17)

which states that the sum of the moments about point O of all the external forces acting on a system of particles is equal to the time rate of change of the total angular momentum of the system about point O. Although O has been chosen here as the origin of coordinates, it actually can represent any fixed point in the inertial frame of reference. z fi

i

Fi

ri y O

x

Inertial coordinate system (b)

Fig. 15–21 (cont.)

15.6

relation Between moment of a forCe and angular momentum

283

15.12

ExaMPlE

The box shown in Fig. 15–22a has a mass m and travels down the smooth circular ramp such that when it is at the angle u it has a speed v. Determine its angular momentum about point O at this instant and the rate of increase in its speed, i.e., at . r sin u

O

O

r u

n

r

u W

v

15

t N (a)

(b)

Fig. 15–22

SolutIon Since v is tangent to the path, applying Eq. 15–12 the angular momentum is Ans.

HO = r mvb

The rate of increase in its speed (dv>dt) can be found by applying Eq. 15–15. From the free-body diagram of the box, Fig. 15–22b, it can be seen that only the weight W = mg contributes a moment about point O. We have # c + M O = HO;

mg(r sin u) =

d (r mv) dt

Since r and m are constant, mgr sin u = r m

dv dt

dv = g sin u dt

Ans.

NOTE: This same result can, of course, be obtained from the equation

of motion applied in the tangential direction, Fig. 15–22b, i.e., + bFt = mat;

mg sin u = m a

dv b dt

dv = g sin u dt

Ans.

284

Chapter 15

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partiCle: impulse

15.7

and

momentum

Principle of Angular Impulse and Momentum

Principle of Angular Impulse and Momentum. If Eq. 15–15 is rewritten in the form MO dt = d HO and integrated, assuming that at time t = t1 , HO = (HO)1 and at time t = t2 , HO = (HO)2, we have



15

Lt1

t2

MO dt = (HO)2 - (HO)1

or

(HO)1 + 

Lt1

t2

MO dt = (HO)2

(15–18)

This equation is referred to as the principle of angular impulse and momentum. The initial and final angular momenta (HO)1 and (HO)2 are defined as the moment of the linear momentum of the particle (HO = r * mv) at the instants t1 and t2 , respectively. The second term on the left side,  1 MO dt, is called the angular impulse. It is determined by integrating, with respect to time, the moments of all the forces acting on the particle over the time period t1 to t2 . Since the moment of a force about point O is MO = r * F, the angular impulse may be expressed in vector form as

angular impulse =

Lt1

t2

MO dt =

Lt1

t2

(r * F) dt

(15–19)

Here r is a position vector which extends from point O to any point on the line of action of F. In a similar manner, using Eq. 15–18, the principle of angular impulse and momentum for a system of particles may be written as

(HO)1 + 

Lt1

t2

MO dt = (HO)2

(15–20)

15.7

prinCiple of angular impulse and momentum

285

Here the first and third terms represent the angular momenta of all the particles [HO = (ri * mvi)] at the instants t1 and t2 . The second term is the sum of the angular impulses given to all the particles from t1 to t2 . Recall that these impulses are created only by the moments of the external forces acting on the system where, for the ith particle, MO = ri * Fi .

Vector Formulation. Using impulse and momentum principles, it is therefore possible to write two equations which define the particle’s motion, namely, Eqs. 15–3 and Eqs. 15–18, restated as

mv1 + g (HO)1 + g

Lt1

Lt1

15

t2

F dt = mv2 (15–21)

t2

MO dt = (HO)2

Scalar Formulation.

In general, the above equations can be expressed in x, y, z component form. If the particle is confined to move in the x–y plane, then three scalar equations can be written to express the motion, namely,

m(vx )1 + g m(vy )1 + g

(HO)1 + 

Lt1

t2

Lt1

Lt1

Fx dt = m(vx )2 t2

Fy dt = m(vy )2

(15–22)

t2

M O dt = (HO)2

The first two of these equations represent the principle of linear impulse and momentum in the x and y directions, which has been discussed in Sec.  15–1, and the third equation represents the principle of angular impulse and momentum about the z axis.

286

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and

momentum

Conservation of Angular Momentum.

y

When the angular impulses acting on a particle are all zero during the time t1 to t2 , Eq. 15–18 reduces to the following simplified form: (HO)1 = (HO)2

F

x

O

Fig. 15–23

15

(15–23)

This equation is known as the conservation of angular momentum. It states that from t1 to t2 the particle’s angular momentum remains constant. Obviously, if no external impulse is applied to the particle, both linear and angular momentum will be conserved. In some cases, however, the particle’s angular momentum will be conserved and linear momentum may not. An example of this occurs when the particle is subjected only to a central force (see Sec. 13.7). As shown in Fig. 15–23, the impulsive central force F is always directed toward point O as the particle moves along the path. Hence, the angular impulse (moment) created by F about the z axis is always zero, and therefore angular momentum of the particle is conserved about this axis. From Eq. 15–20, we can also write the conservation of angular momentum for a system of particles as (HO)1 = (HO)2

(15–24)

In this case the summation must include the angular momenta of all particles in the system.

procedure for analysis When applying the principles of angular impulse and momentum, or the conservation of angular momentum, it is suggested that the following procedure be used. (© Petra Hilke/Fotolia) z

W

Free-Body Diagram. • Draw the particle’s free-body diagram in order to determine any axis about which angular momentum may be conserved. For this to occur, the moments of all the forces (or impulses) must either be parallel or pass through the axis so as to create zero moment throughout the time period t1 to t2 .

• The direction and sense of the particle’s initial and final velocities N

Provided air resistance is neglected, the passengers on this amusement-park ride are subjected to a conservation of angular momentum about the z axis of rotation. As shown on the free-body diagram, the line of action of the normal force N of the seat on the passenger passes through this axis, and the passenger’s weight W is parallel to it. Thus, no angular impulse acts around the z axis.

should also be established.

• An alternative procedure would be to draw the impulse and momentum diagrams for the particle.

Momentum Equations. • Apply the t principle of angular impulse and momentum, (HO)1 +  1t12MOdt = (HO)2 , or if appropriate, the conservation of angular momentum, (HO)1 = (HO)2 .

15.7

ExaMPlE

287

prinCiple of angular impulse and momentum

15.13

The 1.5-Mg car travels along the circular road as shown in Fig. 15–24a. If the traction force of the wheels on the road is F = (150t2) N, where t is in seconds, determine the speed of the car when t = 5 s. The car initially travels with a speed of 5 m>s. Neglect the size of the car.

100 m

15 F

(a)

Free-body Diagram. The free-body diagram of the car is shown in Fig. 15–24b. If we apply the principle of angular impulse and momentum about the z axis, then the angular impulse created by the weight, normal force, and radial frictional force will be eliminated since they act parallel to the axis or pass through it. Principle of Angular Impulse and Momentum. (Hz )1 +  r m c(vc)1 +

Lt1

Lt1

z

t2

M z dt = (Hz )2 W  1500 (9.81)N

t2

r F dt = r m c(vc)2

(100 m)(1500 kg)(5 m>s) +

L0

r  100 m

5s

(100 m)[(150t2) N] dt = (100 m)(1500 kg)(vc)2

750(103) + 5000t3 2

5s 0

Fr

F  (150t2)N N

= 150(103)(vc)2

(vc)2 = 9.17 m>s

Ans.

(b)

Fig. 15–24

288

Chapter 15

ExaMPlE

partiCle: impulse

and

momentum

The 0.8-kg ball B, shown in Fig. 15–25a, is attached to a cord which passes through a hole at A in a smooth table. When the ball is r1 = 0.875 m from the hole, it is rotating around in a circle such that its speed is v1 = 2 m>s. By applying the force F the cord is pulled downward through the hole with a constant speed vc = 3 m>s. Determine (a) the speed of the ball at the instant it is r2 = 0.3 m from the hole, and (b) the amount of work done by F in shortening the radial distance from r1 to r2 . Neglect the size of the ball.

r2

2 m/s

SolutIon part (a) Free-body Diagram. As the ball moves from r1 to r2 , Fig. 15–25b, the cord force F on the ball always passes through the z axis, and the weight and NB are parallel to it. Hence the moments, or angular impulses created by these forces, are all zero about this axis. Therefore, angular momentum is conserved about the z axis. Conservation of angular Momentum. The ball’s velocity v2 is resolved into two components. The radial component, 3 m>s, is known; however, it produces zero angular momentum about the z axis. Thus, H1 = H2

vc  3 m/s

15

of a

15.14

A r1 B

KinetiCs

F

(a)

r1mBv1 = r2mBv2= 0.875 m(0.8 kg) 2 m>s = 0.3 m(0.8 kg)v2=

z

v¿2

3 m/s

v2

v2= = 5.833 m>s

r2  0.3 m

The speed of the ball is thus

0.8 kg

r1  0.875 m

F

2 m/s (b)

Fig. 15–25

NB

v2 = 2(5.833 m>s)2 + (3 m>s)2 = 6.56 m>s

Ans.

part (b). The only force that does work on the ball is F. (The normal force and weight do not move vertically.) The initial and final kinetic energies of the ball can be determined so that from the principle of work and energy we have T 1 + U1 - 2 = T 2 1 1 (0.8 kg)(2 m>s)2 + UF = (0.8 kg)(6.559 m>s)2 2 2 UF = 15.6 J

Ans.

NOTE: The force F is not constant because the normal component of acceleration, an = v2 >r, changes as r changes.

15.7

ExaMPlE

15.15

The 2-kg disk shown in Fig. 15–26a rests on a smooth horizontal surface and is attached to an elastic cord that has a stiffness k c = 20 N>m and is initially unstretched. If the disk is given a velocity (vD)1 = 1.5 m>s, perpendicular to the cord, determine the rate at which the cord is being stretched and the speed of the disk at the instant the cord is stretched 0.2 m.

z

O

15

z y 2(9.81) N

FC

O 0.5 m

(vD= )2 = 1.071 m>s Conservation of energy. The speed of the disk can be obtained by applying the conservation of energy equation at the point where the disk was launched and at the point where the cord is stretched 0.2 m.

+ 21kx 21 = 21m D(vD)22 + 21kx 22

kg)(1.5 m>s)2 + 0 = 21(2 kg)(vD)22 + 21(20 N>m)(0.2 m)2 Ans.

Having determined (vD)2 and its component (vD= )2 , the rate of stretch of the cord, or radial component, (vD> )2 is determined from the Pythagorean theorem, (vD> )2 = 2(vD)22 - (vD= )22

Ans.

ND

x (b)

(v¿¿D)2

0.2 m (v¿D)2

Fig. 15–26

T1 + V 1 = T2 + V 2

= 0.838 m>s

x

(vD)1

0.5 m (2 kg)(1.5 m/s) = 0.7 m(2 kg)(vD= )2

= 2(1.360 m>s)2 - (1.071 m>s)2

0.5 m

0.5 m

r1m D(vD)1 = r2m D(vD= )2

(vD)2 = 1.360 m>s = 1.36 m>s

(vD)1  1.5 m/s

(a)

(HO)1 = (HO)2

1 2 2 m D(vD)1

y kc  20 N/m

SolutIon Free-body Diagram. After the disk has been launched, it slides along the path shown in Fig. 15–26b. By inspection, angular momentum about point O (or the z axis) is conserved, since none of the forces produce an angular impulse about this axis. Also, when the distance is 0.7 m, only the transverse component (vD= )2 produces angular momentum of the disk about O. Conservation of angular Momentum. The component (vD= )2 can be obtained by applying the conservation of angular momentum about O (the z axis).

1 2 (2

289

prinCiple of angular impulse and momentum

(vD)2

290

Chapter 15

KinetiCs

of a

partiCle: impulse

momentum

and

FuNDaMeNtaL prObLeMs F15–19. The 2-kg particle A has the velocity shown. Determine its angular momentum HO about point O. y

10 m/s

5 3

4

F15–22. The 5-kg block is moving around the circular path centered at O on the smooth horizontal plane when it is subjected to the force F = (10t) N, where t is in seconds. If the block starts from rest, determine its speed when t = 4 s. Neglect the size of the block. The force maintains the same constant angle tangent to the path.

2 kg A 3m O

15

O

x

1.5 m

Prob. F15–19

4m

4

A

F15–20. The 2-kg particle A has the velocity shown. Determine its angular momentum HP about point P. 15 m/s

30

y

3

F  (10t) N

5

Prob. F15–22 F15–23. The 2-kg sphere is attached to the light rigid rod, which rotates in the horizontal plane centered at O. If the system is subjected to a couple moment M = (0.9t2) N # m, where t is in seconds, determine the speed of the sphere at the instant t = 5 s starting from rest.

A 2m P 3m

O

M  (0.9t2) Nm

2m

0.6 m

x

Prob. F15–20

F15–21. Initially the 5-kg block is moving with a constant speed of 2 m>s around the circular path centered at O on the smooth horizontal plane. If a constant tangential force F = 5 N is applied to the block, determine its speed when t = 3 s. Neglect the size of the block.

Prob. F15–23 F15–24. Two identical 10-kg spheres are attached to the light rigid rod, which rotates in the horizontal plane centered at pin O. If the spheres are subjected to tangential forces of P = 10 N, and the rod is subjected to a couple moment M = (8t) N # m, where t is in seconds, determine the speed of the spheres at the instant t = 4 s. The system starts from rest. Neglect the size of the spheres. P  10 N 0.5 m

1.5 m

O

M  (8t) Nm

0.5 m O

A F5N

2 m/s

Prob. F15–21

P  10 N Prob. F15–24

15.7

291

prinCiple of angular impulse and momentum

prObLeMs 15–94. Determine the angular momentum HO of each of the particles about point O. 15–95. Determine the angular momentum Hp of each of the particles about point P.

15–98. Determine the angular momentum HO of the 3-kg particle about point O. 15–99. Determine the angular momentum HP of the 3-kg particle about point P.

y 8m

6 kg A

z 3 kg

60

A 6 m/s

4 m/s

P

15

2m

12 m

2m P

2m

O

1.5 m

4 kg

5m

1.5 m

30

B 6m

C

13 12

5

O

2m

6 m/s

2m

2m

1.5 m 1m

x

x

3m

2 kg

B

2.6 m/s

3m

y

Probs. 15–98/99

Probs. 15–94/95 *15–96. Determine the angular momentum Ho of each of the two particles about point O. 15–97. Determine the angular momentum Hp of each of the two particles about point P.

*15–100. Each ball has a negligible size and a mass of 10  kg  and is attached to the end of a rod whose mass may  be  neglected. If the rod is subjected to a torque M = (t2 + 2) N # m, where t is in seconds, determine the speed of each ball when t = 3 s. Each ball has a speed v = 2 m>s when t = 0.

y 4m

P

M  (t 2  2) N  m 5m

1.5 m

4m

O 8 m/s

3 5

4

B 4 kg

2m

3 kg A

Probs. 15–96/97

0.5 m 1m

30

6 m/s

x

v

Prob. 15–100

v

292

Chapter 15

KinetiCs

of a

partiCle: impulse

15–101. The two spheres each have a mass of 3 kg and are attached to the rod of negligible mass. If a torque M = (6e0.2t) N # m, where t is in seconds, is applied to the rod as shown, determine the speed of each of the spheres in 2 s, starting from rest. 15–102. The two spheres each have a mass of 3 kg and are attached to the rod of negligible mass. Determine the time the torque M = (8t) N # m, where t is in seconds, must be applied to the rod so that each sphere attains a speed of 3 m>s starting from rest.

and

momentum

*15–104. A child having a mass of 50 kg holds her legs up as shown as she swings downward from rest at u1 = 30. Her center of mass is located at point G1. When she is at the bottom position u = 0, she suddenly lets her legs come down, shifting her center of mass to position G2. Determine her speed in the upswing due to this sudden movement and the angle u2 to which she swings before momentarily coming to rest. Treat the child’s body as a particle.

3m

15

0.4 m

2.80 m

u2 u1

30° G1

G2

0.4 m

G2

M

Prob. 15–104 Probs. 15–101/102

15–103. If the rod of negligible mass is subjected to a couple moment of M = (30t2) N # m, and the engine of the car supplies a traction force of F = (15t) N to the wheels, where t is in seconds, determine the speed of the car at the instant t = 5 s. The car starts from rest. The total mass of the car and rider is 150 kg. Neglect the size of the car.

15–105. When the 2-kg bob is given a horizontal speed of 1.5 m>s, it begins to move around the horizontal circular path A. If the force F on the cord is increased, the bob rises and then moves around the horizontal circular path B. Determine the speed of the bob around path B. Also, find the work done by force F.

300 mm B

600 mm

4m A M  (30t 2) Nm

F  15t N

Prob. 15–103

F

Prob. 15–105

15.7 15–106. A small particle having a mass m is placed inside the semicircular tube. The particle is placed at the position shown and released. Apply the principle of angular # momentum about point O (M O = HO), and show that the motion of$ the particle is governed by the differential equation u + (g>R) sin u = 0.

293

prinCiple of angular impulse and momentum

*15–108. The two blocks A and B each have a mass of 400 g. The blocks are fixed to the horizontal rods, and their initial velocity along the circular path is 2 m>s. If a couple moment of M = 0.6 N # m is applied about CD of the frame, determine the speed of the blocks when t = 3 s. The mass of the frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks.

O R

u

C

15 M  0.6 N  m

A

B 0.3 m

Prob. 15–106

0.3 m

D

Prob. 15–108 15–107. At the instant r = 1.5 m, the 5-kg disk is given a speed of v = 5 m>s, perpendicular to the elastic cord. Determine the speed of the disk and the rate of shortening of the elastic cord at the instant r = 1.2 m. The disk slides on the smooth horizontal plane. Neglect its size. The cord has an unstretched length of 0.5 m.

A

15–109. The ball B has mass of 10 kg and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque M = (3t2 + 5t + 2) N # m, where t is in seconds, determine the speed of the ball when t = 2 s. The ball has a speed v = 2 m>s when t = 0.

r M

k

B

200 N/m

1.5 m

v

B

5 m/s

v

Prob. 15–107

Prob. 15–109

294

Chapter 15

KinetiCs

of a

partiCle: impulse

15–110. The amusement park ride consists of a 200-kg car and passenger that are traveling at 3 m>s along a circular path having a radius of 8 m. If at t = 0, the cable OA is pulled in toward O at 0.5 m>s, determine the speed of the car when t = 4 s. Also, determine the work done to pull in the cable.

and

momentum

*15–112. A toboggan and rider, having a total mass of 150  kg, enter horizontally tangent to a 90° circular curve with a velocity of vA = 70 km>h. If the track is flat and banked at an angle of 60°, determine the speed vB and the angle u of “descent,” measured from the horizontal in a vertical x–z plane, at which the toboggan exists at B. Neglect friction in the calculation.

z O

A

15

A

r

vA  70 km/ h B

rA  60 m 60 rB  57 m 55 m 55 m 90

60

y

u vB

x

Prob. 15–110

Prob. 15–112

15–111. A small block having a mass of 0.1 kg is given a horizontal velocity v1 = 0.4 m>s of when r1 = 500 mm. It slides along the smooth conical surface. Determine the distance h it must descend for it to reach a speed of v2 = 2 m>s. Also, what is the angle of descent u, that is, the angle measured from the horizontal to the tangent of the path?

15–113. An earth satellite of mass 700 kg is launched into a free-flight trajectory about the earth with an initial speed of vA = 10 km>s when the distance from the center of the earth is rA = 15 Mm. If the launch angle at this position is fA = 70, determine the speed vB of the satellite and its closest distance rB from the center of the earth. The earth has a mass M e = 5.976(1024) kg. Hint: Under these conditions, the satellite is subjected only to the earth’s gravitational force, F = GM em s >r2, Eq. 13–1. For part of the solution, use the conservation of energy.

r1 = 500 mm υ 1 = 0.4 m>s h

vB

r2

θ v2

rB

fA rA vA 30°

Prob. 15–111

Prob. 15–113

15.8

15.8

295

steady flow of a fluid stream

Steady Flow of a Fluid Stream

Up to this point we have restricted our study of impulse and momentum principles to a system of particles contained within a closed volume. In this section, however, we will apply the principle of impulse and momentum to the steady mass flow of fluid particles entering into and then out of a control volume. This volume is defined as a region in space where fluid particles can flow into or out of the region. The size and shape of the control volume is frequently made to coincide with the solid boundaries and openings of a pipe, turbine, or pump. Provided the flow of the fluid into the control volume is equal to the flow out, then the flow can be classified as steady flow.

Principle of Impulse and Momentum. Consider the steady

flow of a fluid stream in Fig. 15–27a that passes through a pipe. The region within the pipe and its openings will be taken as the control volume. As shown, the fluid flows into and out of the control volume with velocities vA and vB , respectively. The change in the direction of the fluid flow within the control volume is caused by an impulse produced by the resultant external force exerted on the control surface by the wall of the pipe. This resultant force can be determined by applying the principle of impulse and momentum to the control volume.

A

15

vA B vB

(a)

Fig. 15–27

The conveyor belt must supply frictional forces to the gravel that falls upon it in order to change the momentum of the gravel stream, so that it begins to travel along the belt.

The air on one side of this fan is essentially at rest, and as it passes through the blades its momentum is increased. To change the momentum of the air flow in this manner, the blades must exert a horizontal thrust on the air stream. As the blades turn faster, the equal but opposite thrust of the air on the blades could overcome the rolling resistance of the wheels on the ground and begin to move the frame of the fan.

296

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

F dt

dm dm vA

mv

mv

A

A

A

B rA

O

dm B

B

r

r¿

Time t

O

r

Time dt

O

dm vB

rB

Time t  dt

15 (b)

As indicated in Fig. 15–27b, a small amount of fluid having a mass dm is about to enter the control volume through opening A with a velocity of vA at time t. Since the flow is considered steady, at time t + dt, the same amount of fluid will leave the control volume through opening B with a velocity vB . The momenta of the fluid entering and leaving the control volume are therefore dm vA and dm vB , respectively. Also, during the time dt, the momentum of the fluid mass within the control volume remains constant and is denoted as mv. As shown on the center diagram, the resultant external force exerted on the control volume produces the impulse F dt. If we apply the principle of linear impulse and momentum, we have dm vA + mv + F dt = dm vB + mv If r, rA , rB are position vectors measured from point O to the geometric centers of the control volume and the openings at A and B, Fig. 15–27b, then the principle of angular impulse and momentum about O becomes

AA dm AB

rA * dm vA + r * mv + r * F dt = r * mv + rB * dm vB

dsA dsB

dm

Dividing both sides of the above two equations by dt and simplifying, we get

(c)

Fig. 15–27 (cont.)

dm (v - vA ) dt B

(15–25)

dm (r * vB - rA * vA ) dt B

(15–26)

F =

MO =

15.8

steady flow of a fluid stream

297

The term dm >dt is called the mass flow. It indicates the constant amount of fluid which flows either into or out of the control volume per unit of time. If the cross-sectional areas and densities of the fluid at the entrance A are A A , rA and at exit B, A B , rB , Fig. 15–27c, then for an incompressible fluid,the continuity of mass requires dm = rdV = rA (dsA A A ) = rB (dsB A B ). Hence, during the time dt, since vA = dsA >dt and vB = dsB >dt, we have dm >dt = rA vA A A = rB vB A B or in general, dm = rvA = rQ dt

(15–27)

The term Q = vA measures the volume of fluid flow per unit of time and is referred to as the discharge or the volumetric flow.

procedure for analysis Problems involving steady flow can be solved using the following procedure. Kinematic Diagram.

• Identify the control volume. If it is moving, a kinematic diagram

may be helpful for determining the entrance and exit velocities of the fluid flowing into and out of its openings since a relativemotion analysis of velocity will be involved.

• The measurement of velocities vA and vB must be made by an observer fixed in an inertial frame of reference.

• Once the velocity of the fluid flowing into the control volume is determined, the mass flow is calculated using Eq. 15–27.

Free-Body Diagram.

• Draw the free-body diagram of the control volume in order to

establish the forces F that act on it. These forces will include the support reactions, the weight of all solid parts and the fluid contained within the control volume, and the static gauge pressure forces of the fluid on the entrance and exit sections.* The gauge pressure is the pressure measured above atmospheric pressure, and so if an opening is exposed to the atmosphere, the gauge pressure there will be zero.

Equations of Steady Flow.

• Apply the equations of steady flow, Eq. 15–25 and 15–26, using the appropriate components of velocity and force shown on the kinematic and free-body diagrams.

* In the SI system, pressure is measured using the pascal (Pa), where 1Pa = 1 N>m2.

15

298

Chapter 15

ExaMPlE

KinetiCs

0.1 m

O G

0.1 m

0.125 m

B vB

0.1 m 0.3 m (a)

15

y

and

momentum

Fy

x

G

O

20(9.81) N 0.125 m 0.3 m

(b)

Fig. 15–28

Determine the components of reaction which the fixed pipe joint at A exerts on the elbow in Fig. 15–28a, if water flowing through the pipe is subjected to a static gauge pressure of 100 kPa at A. The discharge at B is QB = 0.2 m3 >s. Water has a density rw = 1000 kg>m3, and the water-filled elbow has a mass of 20 kg and center of mass at G. SolutIon We will consider the control volume to be the outer surface of the elbow. Using a fixed inertial coordinate system, the velocity of flow at A and B and the mass flow rate can be obtained from Eq. 15–27. Since the density of water is constant, QB = QA = Q. Hence, dm = rwQ = (1000 kg>m3)(0.2 m3 >s) = 200 kg>s dt 0.2 m3 >s Q vB = = = 25.46 m>s T AB p(0.05 m)2 vA =

A MO FA Fx

partiCle: impulse

15.16 A

vA

of a

B

0.2 m3 >s Q = = 6.37 m>s S AA p(0.1 m)2

Free-body Diagram. As shown on the free-body diagram of the control volume (elbow) Fig. 15–28b, the fixed connection at A exerts a resultant couple moment MO and force components Fx and Fy on the elbow. Due to the static pressure of water in the pipe, the pressure force acting on the open control surface at A is FA = pAAA. Since 1 kPa = 1000 N>m2, FA = pA A A = [100(103) N>m2][p(0.1 m)2] = 3141.6 N There is no static pressure acting at B, since the water is discharged at atmospheric pressure; i.e., the pressure measured by a gauge at B is equal to zero, pB = 0. equations of steady Flow. + F = dm (v - v ); -F + 3141.6 N = 200 kg>s(0 - 6.37 m>s) S x Ax x dt Bx Fx = 4.41 kN Ans. dm + c Fy = (v - vA y); -Fy - 20(9.81) N = 200 kg>s(-25.46 m>s - 0) dt By Fy = 4.90 kN Ans. If moments are summed about point O, Fig. 15–28b, then Fx , Fy , and the static pressure FA are eliminated, as well as the moment of momentum of the water entering at A, Fig. 15–28a. Hence, dm c + M O = (d v - dOA vA ) dt OB B M O + 20(9.81) N (0.125 m) = 200 kg>s[(0.3 m)(25.46 m>s) - 0] M O = 1.50 kN # m

Ans.

15.8

ExaMPlE

299

steady flow of a fluid stream

15.17

A 50-mm-diameter water jet having a velocity of 7.5 m>s impinges upon a single moving blade, Fig. 15–29a. If the blade moves with a constant velocity of 1.5 m>s away from the jet, determine the horizontal and vertical components of force which the blade is exerting on the water. What power does the water generate on the blade? Water has a density of rw = 1000 kg>m3. SolutIon Kinematic Diagram. Here the control volume will be the stream of water on the blade. From a fixed inertial coordinate system, Fig. 15–29b, the rate at which water enters the control volume at A is vA = 5 7.5i 6 m>s The relative-flow velocity within the control volume is vw>cv = vw - vcv = 7.5i - 1.5i = 5 6i 6 m>s. Since the control volume is moving with a velocity of vcv = 5 1.5i 6 m>s, the velocity of flow at B measured from the fixed x, y axes is the vector sum, shown in Fig. 15–29b. Here, vB = vcv + vw>cv = 5 1.5i + 6j 6 m>s

Free-body Diagram. The free-body diagram of the control volume is shown in Fig. 15–29c. The weight of the water will be neglected in the calculation, since this force will be small compared to the reactive components Fx and Fy . equations of steady Flow. dm F = (v - vA ) dt B -Fxi + Fyj = 3.75 p(1.5i + 6j - 7.5i) Equating the respective i and j components gives

P = (70.69 N) N(1.5 m>s) = 106 W

vbl  1.5 m/s

A vw  7.5 m/s (a)

15 vB

vw/cv y

B

vcv

vbl

vA A (b) y

x Fx i

Fy j (c)

Ans. Ans.

The water exerts equal but opposite forces on the blade. Since the water force which causes the blade to move forward horizontally with a velocity of 1.5 m>s is Fx = 70.69 N, then from Eq. 14–10 the power is P = F # v;

50 mm

x

Thus, the mass flow of water onto the control volume that undergoes a momentum change is dm 25 2 = rw(vw>cv)AA = (1000) (6) c pa b d = 3.75p kg>s dt 1000

Fx = 3.75p (6) = 70.69 N = 70.7 N d Fy = 3.75p(6) = 70.69 N = 70.7 N c

B

Fig. 15–29

300

Chapter 15

KinetiCs

of a

partiCle: impulse

*15.9

and

momentum

Propulsion with Variable Mass

A Control Volume That Loses Mass.

15

v

Control Volume

ve m me

Consider a device such as a rocket which at an instant of time has a mass m and is moving forward with a velocity v, Fig. 15–30a. At this same instant the amount of mass m e is expelled from the device with a mass flow velocity ve . For the analysis, the control volume will include both the mass m of the device and the expelled mass m e . The impulse and momentum diagrams for the control volume are shown in Fig. 15–30b. During the time dt, its velocity is increased from v to v + dv since an amount of mass dm e has been ejected and thereby gained in the exhaust. This increase in forward velocity, however, does not change the velocity ve of the expelled mass, as seen by a fixed observer, since this mass moves with a constant velocity once it has been ejected. The impulses are created by Fcv, which represents the resultant of all the external forces, such as drag or weight, that act on the control volume in the direction of motion. This force resultant does not include the force which causes the control volume to move forward, since this force (called a thrust) is internal to the control volume; that is, the thrust acts with equal magnitude but opposite direction on the mass m of the device and the expelled exhaust mass m e .* Applying the principle of impulse and momentum to the control volume, Fig. 15–30b, we have + ) (S

(a)

mv - m eve + Fcv dt = (m - dm e)(v + dv) - (m e + dm e)ve

or Fcv dt = -v dm e + m dv - dm e dv - ve dm e

(m  dme) (v  dv)

mv (me  dme)ve

meve

 Fcv dt

m me Time t

Time dt

m  dme (me  dme)

Time t  dt

(b)

Fig. 15–30

*F represents the external resultant force acting on the control volume, which is different from F, the resultant force acting only on the device.

15.9

301

propulsion with variaBle mass

Without loss of accuracy, the third term on the right side may be neglected since it is a “second-order” differential. Dividing by dt gives Fcv = m

dm e dv - (v + ve) dt dt

The velocity of the device as seen by an observer moving with the particles of the ejected mass is vD>e = (v + ve), and so the final result can be written as Fcv

dm e dv = m - vD>e dt dt

Control Volume FD

(15–28)

Here the term dm e >dt represents the rate at which mass is being ejected. To illustrate an application of Eq. 15–28, consider the rocket shown in Fig. 15–31, which has a weight W and is moving upward against an atmospheric drag force FD . The control volume to be considered consists of the mass of the rocket and the mass of ejected gas m e . Applying Eq. 15–28 gives (+ c )

-FD - W =

15

dm e W dv - vD>e g dt dt

v

The last term of this equation represents the thrust T which the engine exhaust exerts on the rocket, Fig. 15–31. Recognizing that dv>dt = a, we can therefore write (+ c )

T - FD - W =

W

W a g

If a free-body diagram of the rocket is drawn, it becomes obvious that this equation represents an application of F = ma for the rocket.

ve

T

A Control Volume That Gains Mass. A device such as a scoop

or a shovel may gain mass as it moves forward. For example, the device shown in Fig. 15–32a has a mass m and moves forward with a velocity v. At this instant, the device is collecting a particle stream of mass m i . The flow velocity vi of this injected mass is constant and independent of the velocity v such that v 7 vi . The control volume to be considered here includes both the mass of the device and the mass of the injected particles. Control Volume v vi m mi

(a)

Fig. 15–32

Fig. 15–31

302

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum (m  dmi) (v  dv)

mv

(mi  dmi) vi

mivi m + dmi

m  Fcv dt

mi Time t

mi  dmi Time t  dt

Time dt (b)

Fig. 15–32 (cont.)

15

a  Fs

m

R

(c)

The impulse and momentum diagrams are shown in Fig. 15–32b. Along with an increase in mass dm i gained by the device, there is an assumed increase in velocity dv during the time interval dt. This increase is caused by the impulse created by Fcv , the resultant of all the external forces acting on the control volume in the direction of motion. The force summation does not include the retarding force of the injected mass acting on the device. Why? Applying the principle of impulse and momentum to the control volume, we have + ) (S

mv + m ivi + Fcv dt = (m + dm i)(v + dv) + (m i - dm i)vi

Using the same procedure as in the previous case, we may write this equation as Fcv = m

dm i dv + (v - vi) dt dt

Since the velocity of the device as seen by an observer moving with the particles of the injected mass is vD>i = (v - vi), the final result can be written as dm i dv (15–29) Fcv = m + vD>i dt dt where dm i >dt is the rate of mass injected into the device. The last term in this equation represents the magnitude of force R, which the injected mass exerts on the device, Fig. 15–32c. Since dv>dt = a, Eq. 15–29 becomes The scraper box behind this tractor represents a device that gains mass. If the tractor maintains a constant velocity v, then dv>dt = 0 and, because the soil is originally at rest, vD>i = v. Applying Eq. 15–29, the horizontal towing force on the scraper box is then T = 0 + v(dm >dt), where dm >dt is the rate of soil accumulated in the box.

Fcv - R = ma This is the application of F = ma. As in the case of steady flow, problems which are solved using Eqs. 15–28 and 15–29 should be accompanied by an identified control volume and the necessary free-body diagram. With this diagram one can then determine Fcv and isolate the force exerted on the device by the particle stream.

15.9

ExaMPlE

303

propulsion with variaBle mass

15.18

The initial combined mass of a rocket and its fuel is m 0 . A total mass m f of fuel is consumed at a constant rate of dm e >dt = c and expelled at a constant speed of u relative to the rocket. Determine the maximum velocity of the rocket, i.e., at the instant the fuel runs out. Neglect the change in the rocket’s weight with altitude and the drag resistance of the air. The rocket is fired vertically from rest. SolutIon Since the rocket loses mass as it moves upward, Eq. 15–28 can be used for the solution. The only external force acting on the control volume consisting of the rocket and a portion of the expelled mass is the weight W, Fig. 15–33. Hence, + c Fcv = m

dm e dv - vD>e ; dt dt

-W = m

dv - uc dt

(1) (© NASA)

The rocket’s velocity is obtained by integrating this equation. At any given instant t during the flight, the mass of the rocket can be expressed as m = m 0 - (dm e >dt)t = m 0 - ct. Since W = mg, Eq.  1 becomes dv -(m 0 - ct)g = (m 0 - ct) - uc dt Separating the variables and integrating, realizing that v = 0 at t = 0, we have L0

v

dv =

L0

t

t

0

= u lna

m0 b - gt m 0 - ct

(2)

Note that liftoff requires the first term on the right to be greater than the second during the initial phase of motion. The time t needed to consume all the fuel is

Hence,

mf = a

Substituting into Eq. 2 yields

dm e b t = ct dt

gm f m0 b m0 - mf c

T

Fig. 15–33

t = m f >c

vmax = u lna

W

uc - gb dt m 0 - ct

a

v = -u ln(m 0 - ct) - gt 2

15

Ans.

304

Chapter 15

ExaMPlE

KinetiCs

of a

partiCle: impulse

and

momentum

15.19 A chain of length l, Fig. 15–34a, has a mass m. Determine the magnitude of force F required to (a) raise the chain with a constant speed vc , starting from rest when y = 0; and (b) lower the chain with a constant speed vc , starting from rest when y = l.

y

15

(a) F

y

y mg( ) l P

SolutIon part (a). As the chain is raised, all the suspended links are given a sudden downward impulse by each added link which is lifted off the ground. Thus, the suspended portion of the chain may be considered as a device which is gaining mass. The control volume to be considered is the length of chain y which is suspended by F at any instant, including the next link which is about to be added but is still at rest, Fig. 15–34b. The forces acting on the control volume exclude the internal forces P and -P, which act between the added link and the suspended portion of the chain. Hence, Fcv = F - mg(y >l). To apply Eq. 15–29, it is also necessary to find the rate at which mass is being added to the system. The velocity vc of the chain is equivalent to vD>i . Why? Since vc is constant, dvc >dt = 0 and dy >dt = vc . Integrating, using the initial condition that y = 0 when t = 0, gives y = vct. Thus, the mass of the control volume at any instant is m cv = m(y >l) = m(vct>l), and therefore the rate at which mass is added to the suspended chain is vc dm i = ma b dt l Applying Eq. 15–29 using this data, we have

P

+ c Fcv = m

(b) F

Hence,

dvc dm i + vD>i dt dt

y vc F - mga b = 0 + vcm a b l l F = (m >l)(gy + v2c )

y y mg( ) l

(c)

Fig. 15–34

Ans.

part (b). When the chain is being lowered, the links which are expelled (given zero velocity) do not impart an impulse to the remaining suspended links. Why? Thus, the control volume in Part (a) will not be considered. Instead, the equation of motion will be used to obtain the solution. At time t the portion of chain still off the floor is y. The free-body diagram for a suspended portion of the chain is shown in Fig. 15–34c. Thus, + c F = ma;

y F - mga b = 0 l

y F = mga b l

Ans.

15.9

propulsion with variaBle mass

305

prObLeMs 15–114. The fire boat discharges two streams of seawater, each at a flow of 0.25 m3>s and with a nozzle velocity of 50  m>s. Determine the tension developed in the anchor chain, needed to secure the boat. The density of seawater is rsv = 1020 kg>m3.

*15–116. The 200-kg boat is powered by the fan which develops a slipstream having a diameter of 0.75 m. If the fan ejects air with a speed of 14 m>s, measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest. Assume that air has a constant density of rw = 1.22 kg>m3 and that the entering air is essentially at rest. Neglect the drag resistance of the water.

30 45

15

0.75 m 60

Prob. 15–116 Prob. 15–114 15–115. The chute is used to divert the flow of water, Q = 0.6 m3 >s. If the water has a cross-sectional area of 0.05  m2, determine the force components at the pin D and roller C necessary for equilibrium. Neglect the weight of the chute and weight of the water on the chute. rw = 1 Mg>m3 .

15–117. The toy sprinkler for children consists of a 0.2-kg cap and a hose that has a mass per length of 30 g>m. Determine the required rate of flow of water through the 5-mm-diameter tube so that the sprinkler will lift 1.5 m from the ground and hover from this position. Neglect the weight of the water in the tube, rw = 1 Mg>m3.

A

0.12 m

C

2m 1.5 m

D

B

1.5 m

Prob. 15–115

Prob. 15–117

306

Chapter 15

KinetiCs

of a

partiCle: impulse

15–118. The bend is connected to the pipe at flanges A and B as shown. If the diameter of the pipe is 0.3 m and it carries a discharge of 1.35 m3 >s, determine the horizontal and vertical components of force reaction and the moment reaction exerted at the fixed base D of the support.The total weight of the bend and the water within it is 2500 N, with a mass center at point G. The gauge pressure of the water at the flanges at A and B are 120 kN>m2 and 96 kN>m2, respectively. Assume that no force is transferred to the flanges at A and B. The specific weight of water is gv = 10 kN>m3.

and

momentum

*15–120. The hemispherical bowl of mass m is held in equilibrium by the vertical jet of water discharged through a nozzle of diameter d. If the discharge of the water through the nozzle is Q, determine the height h at which the bowl is suspended. The water density is rw. Neglect the weight of the water jet.

h

15

G

0.45 m A

B

Prob. 15–120 45

1.2 m D

Prob. 15–118

15–121. A snowblower having a scoop S with a crosssectional area of As = 0.12 m3 is pushed into snow with a speed of vs = 0.5 m>s. The machine discharges the snow through a tube T that has a cross-sectional area of AT = 0.03 m2 and is directed 60° from the horizontal. If the density of snow is rs = 104 kg>m3, determine the horizontal force P required to push the blower forward, and the resultant frictional force F of the wheels on the ground, necessary to prevent the blower from moving sideways. The wheels roll freely. P

15–119. Water is discharged at 16 m>s against the fixed cone diffuser. If the opening diameter of the nozzle is 40 mm, determine the horizontal force exerted by the water on the diffuser, rw = 1 Mg>m3.

T

60

S 16 m/s

vx

30

Prob. 15–119

Prob. 15–121

F

15.9 15–122. The gauge pressure of water at A is 150.5 kPa. Water flows through the pipe at A with a velocity of 18 m>s, and out the pipe at B and C with the same velocity v. Determine the horizontal and vertical components of force  exerted on the elbow necessary to hold the pipe assembly in  equilibrium. Neglect the weight of water within the pipe and the weight of the pipe. The pipe has a diameter of  50  mm at A, and at B and C the diameter is 30 mm. rw = 1000 kg>m3.

15–125. Water is flowing from the 150-mm-diameter fire hydrant with a velocity vB = 15 m>s. Determine the horizontal and vertical components of force and the moment developed at the base joint A, if the static (gauge) pressure at A is 50 kPa. The diameter of the fire hydrant at A is 200 mm. rw = 1 Mg>m3.

B

v

vB 5

v

307

propulsion with variaBle mass

15 m/s

15

C 3

A

4

B 18 m/s

500 mm

Prob. 15–122

A

15–123. A scoop in front of the tractor collects snow at a rate of 200 kg>s. Determine the resultant traction force T that must be developed on all the wheels as it moves forward on level ground at a constant speed of 5 km>h. The tractor has a mass of 5 Mg. *15–124. The boat has a mass of 180 kg and is traveling forward on a river with a constant velocity of 70 km>h, measured relative to the river. The river is flowing in the opposite direction at 5 km>h. If a tube is placed in the water, as shown, and it collects 40 kg of water in the boat in 80 s, determine the horizontal thrust T on the tube that is required to overcome the resistance due to the water collection and yet maintain the constant speed of the boat. rw = 1 Mg>m3.

Prob. 15–125

15–126. Water is discharged from a nozzle with a velocity of 12 m>s and strikes the blade mounted on the 20-kg cart. Determine the tension developed in the cord, needed to hold the cart stationary, and the normal reaction of the wheels on the cart. The nozzle has a diameter of 50 mm and the density of water is rw = 1000 kg>m3.

45

T

Prob. 15–124

vR  5 km/h

A

B

Prob. 15–126

308

Chapter 15

KinetiCs

of a

partiCle: impulse

15–127. When operating, the air-jet fan discharges air with a speed of vB = 20 m>s into a slipstream having a diameter of 0.5 m. If air has a density of 1.22 kg>m3, determine the horizontal and vertical components of reaction at C and the vertical reaction at each of the two wheels, D, when the fan is in operation. The fan and motor have a mass of 20 kg and a center of mass at G. Neglect the weight of the frame. Due to symmetry, both of the wheels support an equal load. Assume the air entering the fan at A is essentially at rest.

and

momentum

15–129. Each of the two stages A and B of the rocket has a mass of 2 Mg when their fuel tanks are empty. They each carry 500 kg of fuel and are capable of consuming it at a rate of 50 kg>s and eject it with a constant velocity of 2500 m>s, measured with respect to the rocket. The rocket is launched vertically from rest by first igniting stage B. Then stage A is ignited immediately after all the fuel in B is consumed and A has separated from B. Determine the maximum velocity of stage A. Neglect drag resistance and the variation of the rocket’s weight with altitude.

0.5 m A

0.25 m G

15

vB

A

1.5 m

B

D C 0.8 m

0.25 m

Prob. 15–127 Prob. 15–129 *15–128. Sand is discharged from the silo at A at a rate of 50 kg>s with a vertical velocity of 10 m>s onto the conveyor belt, which is moving with a constant velocity of 1.5 m>s. If the conveyor system and the sand on it have a total mass of 750 kg and center of mass at point G, determine the horizontal and vertical components of reaction at the pin support B and roller support A. Neglect the thickness of the conveyor. B 1.5 m/s

15–130. Sand is deposited from a chute onto a conveyor belt which is moving at 0.5 m>s. If the sand is assumed to fall vertically onto the belt at A at the rate of 4 kg>s, determine the belt tension FB to the right of A. The belt is free to move over the conveyor rollers and its tension to the left of A is FC = 400 N.

G

A

30

0.5 m/s

10 m/s 4m

4m

Prob. 15–128

FC = 400 N

A

Prob. 15–130

FB

15.9 15–131. The water flow enters below the hydrant at C at the rate of 0.75 m3 >s. It is then divided equally between the two outlets at A and B. If the gauge pressure at C is 300 kPa, determine the horizontal and vertical force reactions and the moment reaction on the fixed support at C. The diameter of the two outlets at A and B is 75 mm, and the diameter of  the inlet pipe at C is 150 mm. The density of water is rw = 1000 kg>m3. Neglect the mass of the contained water and the hydrant.

propulsion with variaBle mass

309

15–133. Sand drops onto the 2-Mg empty rail car at 50 kg>s from a conveyor belt. If the car is initially coasting at 4 m>s, determine the speed of the car as a function of time.

4 m/s

Prob. 15–133 15–134. The tractor together with the empty tank has a total mass of 4 Mg. The tank is filled with 2 Mg of water. The water is discharged at a constant rate of 50 kg>s with a 15 constant velocity of 5 m>s, measured relative to the tractor. If the tractor starts from rest, and the rear wheels provide a resultant traction force of 250 N, determine the velocity and  acceleration of the tractor at the instant the tank becomes empty.

250 mm

30 A

B

650 mm

600 mm

F C

Prob. 15–134

Prob. 15–131 *15–132. The nozzle has a diameter of 40 mm. If it discharges water uniformly with a downward velocity of 20 m>s against the fixed blade, determine the vertical force exerted by the water on the blade. rw = 1 Mg>m3.

15–135. The earthmover initially carries 10 m3 of sand having a density of 1520 kg>m3. The sand is unloaded horizontally through a 2.5 m3 dumping port P at a rate of 900 kg>s measured relative to the port. Determine the  resultant tractive force F at its front wheels if the acceleration of the earthmover is 0.1 m>s2 when half the sand is dumped. When empty, the earthmover has a mass of 30 Mg. Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll.

40 mm

P 45

45

Prob. 15–132

F

Prob. 15–135

310

Chapter 15

KinetiCs

of a

partiCle: impulse

*15–136. A power lawn mower hovers very close over the ground. This is done by drawing air in at a speed of 6 m>s through an intake unit A, which has a cross-sectional area of A A = 0.25 m2, and then discharging it at the ground, B, where the cross-sectional area is A B = 0.35 m2. If air at A is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has a mass of 15 kg with center of mass at G. Assume that air has a constant density of ra = 1.22 kg/m3.

and

momentum

15–138. The rocket has an initial mass m0, including the fuel. For practical reasons desired for the crew, it is required that it maintain a constant upward acceleration a0. If the fuel is expelled from the rocket at a relative speed ye>r, determine the rate at which the fuel should be consumed to maintain the motion. Neglect air resistance, and assume that the gravitational acceleration is constant.

a0

15

vA A G

Prob. 15–138

B

Prob. 15–136

15–137. The rocket car has a mass of 2 Mg (empty) and carries 120 kg of fuel. If the fuel is consumed at a constant rate of 6 kg>s and ejected from the car with a relative velocity of 800 m>s, determine the maximum speed attained by the car starting from rest. The drag resistance due to the atmosphere is FD = (6.8v2) N, where v is the speed in m>s.

15–139. The 12-Mg jet airplane has a constant speed of 950 km>h when it is flying along a horizontal straight line. Air enters the intake scoops S at the rate of 50 m3 >s. If the engine burns fuel at the rate of 0.4 kg>s and the gas (air and fuel) is exhausted relative to the plane with a speed of 450 m>s, determine the resultant drag force exerted on the plane by air resistance. Assume that air has a constant density of 1.22 kg>m3. Hint: Since mass both enters and exits the plane, Eqs. 15–28 and 15–29 must be combined to yield Fs = m

v

dme dmi dv - vD>e + vD>i . dt dt dt

v  950 km/h

S

Prob. 15–137

Prob. 15–139

15.9 *15–140. The jet is traveling at a speed of 720 km>h. If the fuel is being spent at 0.8 kg>s, and the engine takes in air at 200 kg>s, whereas the exhaust gas (air and fuel) has a relative speed of 12 000 m>s, determine the acceleration of the plane at this instant. The drag resistance of the air is FD = (55v2), where the speed is measured in m>s. The jet has a mass of 7 Mg.

propulsion with variaBle mass

311

15–143. A four-engine commercial jumbo jet is cruising at a constant speed of 800 km>h in level flight when all four engines are in operation. Each of the engines is capable of discharging combustion gases with a velocity of 775 m>s relative to the plane. If during a test two of the engines, one on each side of the plane, are shut off, determine the new cruising speed of the jet. Assume that air resistance (drag) is proportional to the square of the speed, that is, FD = cv2, where c is a constant to be determined. Neglect the loss of mass due to fuel consumption.

720 km/h

15 Prob. 15–140

Prob. 15–143

15–141. The rope has a mass m per unit length. If the end length y = h is draped off the edge of the table, and released, determine the velocity of its end A for any position y, as the rope uncoils and begins to fall.

yh A

*15–144. A commercial jet aircraft has a mass of 150 Mg and is cruising at a constant speed of 850 km>h in level flight (u = 0). If each of the two engines draws in air at a rate of 1000 kg>s and ejects it with a velocity of 900 m>s, relative to the aircraft, determine the maximum angle of inclination u at which the aircraft can fly with a constant speed of 750 km>h. Assume that air resistance (drag) is proportional to the square of the speed, that is, FD = cy2, where c is a constant to be determined. The engines are operating with the same power in both cases. Neglect the amount of fuel consumed.

Prob. 15–141

15–142. A coil of heavy open chain is used to reduce the stopping distance of a sled that has a mass M and travels at a speed of v0. Determine the required mass per unit length of the chain needed to slow down the sled to (1/2)v0 within a distance x = s if the sled is hooked to the chain at x = 0. Neglect friction between the chain and the ground.

u

Prob. 15–144

312

Chapter 15

KinetiCs

of a

partiCle: impulse

15–145. A coil of heavy open chain is used to reduce the stopping distance of a sled that has a mass M and travels at a speed of v0. Determine the required mass per unit length of the chain needed to slow down the sled to (1/4)v0 within a distance x = s if the sled is hooked to the chain at x = 0. Neglect friction between the chain and the ground. 15–146. Determine the magnitude of force F as a function of time, which must be applied to the end of the cord at A to raise the hook H with a constant speed v = 0.4 m>s. Initially the chain is at rest on the ground. Neglect the mass of the cord and the hook. The chain has a mass of 2 kg>m.

and

momentum

*15–148. The truck has a mass of 50 Mg when empty. When it is unloading 5 m3 of sand at a constant rate of 0.8 m3 >s, the sand flows out the back at a speed of 7 m>s, measured relative to the truck, in the direction shown. If the truck is free to roll, determine its initial acceleration just as the load begins to empty. Neglect the mass of the wheels and any frictional resistance to motion. The density of sand is rs = 1520 kg>m3.

A

15 a

H v  0.4 m/s

45 7 m/s

Prob. 15–148 Prob. 15–146 15–147. The 10-Mg helicopter carries a bucket containing 500 kg of water, which is used to fight fires. If it hovers over the land in a fixed position and then releases 50 kg>s of water at 10 m>s, measured relative to the helicopter, determine the initial upward acceleration the helicopter experiences as the water is being released. a

15–149. The car has a mass m0 and is used to tow the smooth chain having a total length l and a mass per unit of length m . If the chain is originally piled up, determine the tractive force F that must be supplied by the rear wheels of the car, necessary to maintain a constant speed v while the chain is being drawn out.

v

F

Prob. 15–147

Prob. 15–149

15.9

propulsion with variaBle mass

313

CONCeptuaL prObLeMs C15–1. The ball travels to the left when it is struck by the bat. If the ball then moves horizontally to the right, determine which measurements you could make in order to determine the net impulse given to the ball. Use numerical values to give an example of how this can be done.

C15–3. The train engine on the left, A, is at rest, and the one on the right, B, is coasting to the left. If the engines are identical, use numerical values to show how to determine the maximum compression in each of the spring bumpers that are mounted in the front of the engines. Each engine is free to roll.

A

Prob. C15–1 C15–2. The steel wrecking “ball” is suspended from the boom using an old rubber tire A. The crane operator lifts the ball then allows it to drop freely to break up the concrete. Explain, using appropriate numerical data, why it is a good idea to use the rubber tire for this work.

B

Prob. C15–3 C15–4. Three train cars each have the same mass and are rolling freely when they strike the fixed bumper. Legs AB and BC on the bumper are pin connected at their ends and the angle BAC is 30 and BCA is 60. Compare the average impulse in each leg needed to stop the motion if the cars have no bumper and if the cars have a spring bumper. Use appropriate numerical values to explain your answer.

A

B

C A

Prob. C15–2

Prob. C15–4

15

314

Chapter 15

KinetiCs

of a

partiCle: impulse

and

momentum

Chapter review Impulse

F

An impulse is defined as the product of force and time. Graphically it represents the area under the F9t diagram. If  the force is constant, then the impulse becomes I = Fc(t2 - t1).

t

I  t 2 F(t)dt 1

t1

15

t2

t

Principle of Impulse and Momentum When the equation of motion, F = ma, and the kinematic equation, a = dv>dt, are combined, we obtain the principle of impulse and momentum. This is a vector equation that can be resolved into rectangular components and used to solve problems that involve force, velocity, and time. For application, the free-body diagram should be drawn in order to account for all the impulses that act on the particle.

mv1 + 

Lt1

t2

F dt = mv2

Conservation of Linear Momentum If the principle of impulse and momentum is applied to a system of particles, then the collisions between the particles produce internal impulses that are equal, opposite, and collinear, and therefore cancel from the equation. Furthermore, if an external impulse is small, that is, the force is small and the time is short, then the impulse can be classified as nonimpulsive and can be neglected. Consequently, momentum for the system of particles is conserved.

m i(vi)1 = m i(vi)2

The conservation-of-momentum equation is useful for finding the final velocity of a particle when internal impulses are exerted between two particles and the initial velocities of the particles is known. If the internal impulse is to be determined, then one of the particles is isolated and the principle of impulse and momentum is applied to this particle.

Impact When two particles A and B have a direct impact, the internal impulse between them is equal, opposite, and collinear. Consequently the conservation of momentum for this system applies along the line of impact.

Plane of contact vA

A

B

vB Line of impact

m A (vA )1 + m B (vB )1 = m A (vA )2 + m B (vB )2

Chapter review

If the final velocities are unknown, a second equation is needed for solution. We must use the coefficient of restitution, e. This experimentally determined coefficient depends upon the physical properties of the colliding particles. It can be expressed as the ratio of their relative velocity after collision to their relative velocity before collision. If the collision is elastic, no energy is lost and e = 1. For a plastic collision e = 0.

(vB )2 - (vA )2 (vA )1 - (vB )1

e =

If the impact is oblique, then the conservation of momentum for the system and the coefficient-ofrestitution equation apply along the line of impact. Also, conservation of momentum for each particle applies perpendicular to this line (plane of contact) because no impulse acts on the particles in this direction.

Plane of contact A

B

Line of impact

u

f

vA

vB

15

Principle of Angular Impulse and Momentum The moment of the linear momentum about an axis (z) is called the angular momentum. The principle of angular impulse and momentum is often used to eliminate unknown impulses by summing the moments about an axis through which the lines of action of these impulses produce no moment. For this reason, a free-body diagram should accompany the solution.

z

(HO)z = (d)(mv) (HO)1 + 

Lt1

t2

O

MO dt = (HO)2

HO

y

d mv

x

Steady Fluid Streams Impulse-and-momentum methods are often used to determine the forces that a device exerts on the mass flow of a fluid—liquid or gas. To do so, a free-body diagram of the fluid mass in contact with the device is drawn in order to identify these forces. Also, the velocity of the fluid as it flows into and out of a control volume for the device is calculated. The equations of steady flow involve summing the forces and the moments to determine these reactions.

F = MO =

dm (v - vA ) dt B

dm (r * vB - rA * vA) dt B

Propulsion with Variable Mass Some devices, such as a rocket, lose mass as they are propelled forward. Others gain mass, such as a shovel. We can account for this mass loss or gain by applying the principle of impulse and momentum to a control volume for the device. From this equation, the force exerted on the device by the mass flow can then be determined.

315

Fcv = m

dm e dv - vD>e dt dt

Loses Mass Fcv = m

dm i dv + vD>i dt dt

Gains Mass

316

Chapter 15

KinetiCs

of a

partiCle: impulse

momentum

and

FuNDaMeNtaL review prObLeMs R15–1. Packages having a mass of 6 kg slide down a smooth chute and land horizontally with a speed of 3 m>s on the surface of a conveyor belt. If the coefficient of kinetic friction between the belt and a package is μk = 0.2, determine the time needed to bring the package to rest on the belt if the belt is moving in the same direction as the package with a speed v = 1 m>s.

R15–3. A 20-kg block is originally at rest on a horizontal surface for which the coefficient of static friction is μs = 0.6 and the coefficient of kinetic friction is μk = 0.5. If a horizontal force F is applied such that it varies with time as shown, determine the speed of the block in 10 s. Hint: First determine the time needed to overcome friction and start the block moving.

15 F 3 m/s

F (N) 1 m/s 200

Prob. R15–1

5

t (s)

10

Prob. R15–3 R15–2. The 50-kg block is hoisted up the incline using the cable and motor arrangement shown. The coefficient of kinetic friction between the block and the surface is mk = 0.4. If the block is initially moving up the plane at v0 = 2 m>s, and at this instant (t = 0) the motor develops a tension in the cord of T = (300 + 1201t) N, where t is in seconds, determine the velocity of the block when t = 2 s.

R15–4. The three freight cars A, B, and C have masses of 10 Mg, 5 Mg, and 20 Mg, respectively. They are traveling along the track with the velocities shown. Car A collides with car B first, followed by car C. If the three cars couple together after collision, determine the common velocity of the cars after the two collisions have taken place.

v0  2 m/s 20 km/h A

5 km/h B

30

Prob. R15–2

Prob. R15–4

25 km/h C

317

fundamental review proBlems R15–5. The 200-g projectile is fired with a velocity of 900  m>s towards the center of the 15-kg wooden block, which rests on a rough surface. If the projectile penetrates and emerges from the block with a velocity of 300 m>s, determine the velocity of the block just after the projectile emerges. How long does the block slide on the rough surface, after the projectile emerges, before it comes to rest again? The coefficient of kinetic friction between the surface and the block is mk = 0.2.

R15–7. Two smooth billiard balls A and B have an equal mass of m = 200 g. If A strikes B with a velocity of (vA)1 =  2 m>s as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e = 0.75.

y A 900 m/s

(vA)1

15

40 x

B Before

Prob. R15–7

300 m/s

After

Prob. R15–5

R15–8. The small cylinder C has a mass of 10 kg and is attached to the end of a rod whose mass may be neglected. If the frame is subjected to a couple M = (8t2 + 5) N # m, where t is in seconds, and the cylinder is subjected to a force of 60 N, which is always directed as shown, determine the speed of the cylinder when t = 2 s. The cylinder has a speed v0 = 2 m/s when t = 0.

R15–6. Block A has a mass of 3 kg and is sliding on a rough horizontal surface with a velocity (vA )1 = 2 m>s when it makes a direct collision with block B, which has a mass of 2 kg and is originally at rest. If the collision is perfectly elastic (e = 1), determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is mk = 0.3.

z 60 N

5 4

0.75 m

x (vA)1 A

3

C M  (8t2  5) Nm

B

Prob. R15–6

Prob. R15–8

v y

Chapter

16

(© TFoxFoto/Shutterstock) Kinematics is important for the design of the mechanism used on this dump truck.

Planar Kinematics of a Rigid Body Chapter ObjeCtives • To classify the various types of rigid-body planar motion. • To investigate rigid-body translation and angular motion about a fixed axis. • To study planar motion using an absolute motion analysis. • To provide a relative motion analysis of velocity and acceleration using a translating frame of reference. • To show how to find the instantaneous center of zero velocity and determine the velocity of a point on a body using this method. • To provide a relative-motion analysis of velocity and acceleration using a rotating frame of reference.

16.1

Planar Rigid-Body Motion

In this chapter, the planar kinematics of a rigid body will be discussed. This study is important for the design of gears, cams, and mechanisms used for many mechanical operations. Once the kinematics is thoroughly understood, then we can apply the equations of motion, which relate the forces on the body to the body’s motion. The planar motion of a body occurs when all the particles of a rigid body move along paths which are equidistant from a fixed plane. There are three types of rigid-body planar motion. In order of increasing complexity, they are

Video Solutions are available for selected questions in this chapter.

320

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

Path of curvilinear translation (b)

Path of rectilinear translation (a)

Rotation about a fixed axis (c)

16

General plane motion (d)

Fig. 16–1

• Translation. This type of motion occurs when a line in the body

• •

remains parallel to its original orientation throughout the motion. When the paths of motion for any two points on the body are parallel lines, the motion is called rectilinear translation, Fig. 16–1a. If the paths of motion are along curved lines, the motion is called curvilinear translation, Fig. 16–1b. Rotation about a fixed axis. When a rigid body rotates about a fixed axis, all the particles of the body, except those which lie on the axis of rotation, move along circular paths, Fig. 16–1c. General plane motion. When a body is subjected to general plane motion, it undergoes a combination of translation and rotation, Fig. 16–1d. The translation occurs within a reference plane, and the rotation occurs about an axis perpendicular to the reference plane.

In the following sections we will consider each of these motions in detail. Examples of bodies undergoing these motions are shown in Fig. 16–2. Curvilinear translation

General plane motion

r

r

Rectilinear translation

Rotation about a fixed axis

Fig. 16–2

16.2

16.2

translation

321

Translation

Consider a rigid body which is subjected to either rectilinear or curvilinear translation in the x–y plane, Fig. 16–3. y¿

y B rB/A rB A rA O

Fixed coordinate system

x¿ Translating coordinate system

x

16 Fig. 16–3

Position.

The locations of points A and B on the body are defined with respect to fixed x, y reference frame using position vectors rA and rB . The translating x , y  coordinate system is fixed in the body and has its origin at A, hereafter referred to as the base point. The position of B with respect to A is denoted by the relative-position vector rB>A (“r of B with respect to A”). By vector addition, rB = rA + rB>A

Velocity.

A relation between the instantaneous velocities of A and B is obtained by taking the time derivative of this equation, which yields vB = vA + drB>A >dt. Here vA and vB denote absolute velocities since these vectors are measured with respect to the x, y axes. The term drB>A >dt = 0, since the magnitude of rB>A is constant by definition of a rigid body, and because the body is translating the direction of rB>A is also constant. Therefore, vB = vA

Acceleration. Taking the time derivative of the velocity equation yields a similar relationship between the instantaneous accelerations of A and B: aB = aA The above two equations indicate that all points in a rigid body subjected to either rectilinear or curvilinear translation move with the same velocity and acceleration. As a result, the kinematics of particle motion, discussed in Chapter 12, can also be used to specify the kinematics of points located in a translating rigid body.

Riders on this Ferris wheel are subjected to curvilinear translation, since the gondolas move in a circular path, yet it always remains in the upright position.

322

Chapter 16

p l a n a r K i n e m at i C s

of a

16.3

rigid Body

Rotation about a Fixed Axis

When a body rotates about a fixed axis, any point P located in the body travels along a circular path. To study this motion it is first necessary to discuss the angular motion of the body about the axis.

dU

Angular Motion. Since a point is without dimension, it cannot

A

have angular motion. Only lines or bodies undergo angular motion. For example, consider the body shown in Fig. 16–4a and the angular motion of a radial line r located within the shaded plane.

V

Angular Position.

At the instant shown, the angular position of r is defined by the angle u, measured from a fixed reference line to r.

Angular Displacement. The change in the angular position, which

O r

16

P du

u

can be measured as a differential dU, is called the angular displacement.* This vector has a magnitude of dU, measured in degrees, radians, or revolutions, where 1 rev = 2p rad. Since motion is about a fixed axis, the direction of dU is always along this axis. Specifically, the direction is determined by the right-hand rule; that is, the fingers of the right hand are curled with the sense of rotation, so that in this case the thumb, or dU, points upward, Fig. 16–4a. In two dimensions, as shown by the top view of the shaded plane, Fig. 16–4b, both u and du are counterclockwise, and so the thumb points outward from the page.

Angular Velocity. The time rate of change in the angular position

is called the angular velocity V (omega). Since dU occurs during an instant of time dt, then,

(a)

(a+)

V

A

O r P

dU

v =

du dt

(16–1)

This vector has a magnitude which is often measured in rad>s. It is expressed here in scalar form since its direction is also along the axis of rotation, Fig. 16–4a. When indicating the angular motion in the shaded plane, Fig. 16–4b, we can refer to the sense of rotation as clockwise or counterclockwise. Here we have arbitrarily chosen counterclockwise rotations as positive and indicated this by the curl shown in parentheses next to Eq. 16–1. Realize, however, that the directional sense of V is actually outward from the page.

u (b)

Fig. 16–4

*It is shown in Sec. 20.1 that finite rotations or finite angular displacements are not vector quantities, although differential rotations dU are vectors.

16.3

rotation aBout a fixed axis

323

Angular Acceleration. The angular acceleration A (alpha) measures the time rate of change of the angular velocity. The magnitude of this vector is

(a+)

a =

dv dt

(16–2)

Using Eq. 16–1, it is also possible to express a as (a+)

a =

d2u dt2

(16–3)

The line of action of A is the same as that for V, Fig. 16–4a; however, its sense of direction depends on whether V is increasing or decreasing. If V is decreasing, then A is called an angular deceleration and therefore has a sense of direction which is opposite to V. By eliminating dt from Eqs. 16–1 and 16–2, we obtain a differential relation between the angular acceleration, angular velocity, and angular displacement, namely, (a+)

a du = v dv

(16–4)

The similarity between the differential relations for angular motion and those developed for rectilinear motion of a particle (v = ds>dt, a = dv>dt, and a ds = v dv) should be apparent.

Constant Angular Acceleration. If the angular acceleration of the body is constant, A = Ac , then Eqs. 16–1, 16–2, and 16–4, when integrated, yield a set of formulas which relate the body’s angular velocity, angular position, and time. These equations are similar to Eqs. 12–4 to 12–6 used for rectilinear motion. The results are (a+) (a+) (a+)

v = v0 + act u = u0 + v0t + 21 act2 v2 = v20 + 2ac(u - u0) Constant Angular Acceleration

(16–5) (16–6) (16–7)

Here u0 and v0 are the initial values of the body’s angular position and angular velocity, respectively.

The gears used in the operation of a crane all rotate about fixed axes. Engineers must be able to relate their angular motions in order to properly design this gear system.

16

324

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

O

r

v

V P O r

dU f P ds rP

v

(d)

Motion of Point P. As the rigid body in Fig. 16–4c rotates, point P travels along a circular path of radius r with center at point O. This path is contained within the shaded plane shown in top view, Fig. 16–4d. (c)

16

Fig. 16–4 (cont.)

Position and Displacement. The position of P is defined by the position vector r, which extends from O to P. If the body rotates du then P will displace ds = r du.

Velocity. The velocity of P has a magnitude which can be found by dividing ds = r du by dt so that v = vr

(16–8)

As shown in Figs. 16–4c and 16–4d, the direction of v is tangent to the circular path. Both the magnitude and direction of v can also be accounted for by using the cross product of V and rP (see Appendix B). Here, rP is directed from any point on the axis of rotation to point P, Fig. 16–4c. We have v = V * rP

(16–9)

The order of the vectors in this formulation is important, since the cross product is not commutative, i.e., V * rP  rP * V. Notice in Fig. 16–4c how the correct direction of v is established by the right-hand rule. The fingers of the right hand are curled from V toward rP (V “cross” rP). The thumb indicates the correct direction of v, which is tangent to the path in the direction of motion. From Eq. B–8, the magnitude of v in Eq. 16–9 is v = vrP sin f, and since r = rP sin f, Fig. 16–4c, then v = vr, which agrees with Eq. 16–8. As a special case, the position vector r can be chosen for rP . Here r lies in the plane of motion and again the velocity of point P is v = V * r

(16–10)

16.3

Acceleration. The acceleration of P can be expressed in terms of its normal and tangential components. Applying Eq. 12–19 and Eq. 12–20, at = dv>dt and an = v2 >r, where r = r, v = vr, and a = dv>dt, we get at = ar

V, A

(16–11)

O an

an = v2r

drP dv dV = * rP + V * dt dt dt

(e)

16

a

O

at P

(16–13)

From the definition of the cross product, the first term on the right has a magnitude at = arP sin f = ar, and by the right-hand rule, A * rP is in the direction of at , Fig. 16–4e. Likewise, the second term has a magnitude an = v2rP sin f = v2r, and applying the right-hand rule twice, first to determine the result vP = V * rP then V * vP , it can be seen that this result is in the same direction as an, shown in Fig. 16–4e. Noting that this is also the same direction as -r, which lies in the plane of motion, we can express an in a much simpler form as an = -v2r. Hence, Eq. 16–13 can be identified by its two components as a = at + an = A * r - v2r

at

an

Recalling that A = dV>dt, and using Eq. 16–9 (drP >dt = v = V * rP), yields a = A * rP + V * (V * rP)

a

r f P rP

(16–12)

The tangential component of acceleration, Figs. 16–4e and 16–4f, represents the time rate of change in the velocity’s magnitude. If the speed of P is increasing, then a t acts in the same direction as v; if the speed is decreasing, a t acts in the opposite direction of v; and finally, if the speed is constant, a t is zero. The normal component of acceleration represents the time rate of change in the velocity’s direction. The direction of an is always toward O, the center of the circular path, Figs. 16–4e and 16–4f. Like the velocity, the acceleration of point P can be expressed in terms of the vector cross product. Taking the time derivative of Eq. 16–9 we have a=

325

rotation aBout a fixed axis

(16–14)

Since at and an are perpendicular to one another, if needed the magnitude of acceleration can be determined from the Pythagorean theorem; namely, a = 2a2n + a2t , Fig. 16–4f.

(f)

Fig. 16–4 (cont.)

326

Chapter 16

rB

p l a n a r K i n e m at i C s

rigid Body

If two rotating bodies contact one another, then the points in contact move along different circular paths, and the velocity and tangential components of acceleration of the points will be the same: however, the normal components of acceleration will not be the same. For example, consider the two meshed gears in Fig. 16–5a. Point A is located on gear B and a coincident point A¿ is located on gear C. Due to the rotational motion, vA = vA¿, Fig. 16–5b, and as a result, vBrB = vCrC or vB = vC (rC >rB ). Also, from Fig. 16–5c, (aA)t = (aA¿)t, so that aB = aC (rC >rB ); however, since both points follow different circular paths, (aA)n Z (aA¿)n and therefore, as shown, aA Z aA¿.

rC A

of a

A¿ C

B (a)

(aA¿)n rB rB A

A¿

rC

(aA)n

A

A¿

rC C

B

C B

16

vA

vA¿

aA (a ) A t (c)

(b)

(aA¿)t aA¿

Fig. 16–5

important points • A body can undergo two types of translation. During rectilinear

translation all points follow parallel straight-line paths, and during curvilinear translation the points follow curved paths that are the same shape.

• All the points on a translating body move with the same velocity and acceleration.

• Points located on a body that rotates about a fixed axis follow circular paths.

• The relation a du = v dv is derived from a = dv>dt and v = du>dt by eliminating dt.

• Once angular motions v and a are known, the velocity and acceleration of any point on the body can be determined.

• The velocity always acts tangent to the path of motion. • The acceleration has two components. The tangential acceleration

measures the rate of change in the magnitude of the velocity and can be determined from at = ar. The normal acceleration measures the rate of change in the direction of the velocity and can be determined from an = v2r.

16.3

rotation aBout a fixed axis

327

procedure for analysis The velocity and acceleration of a point located on a rigid body that is rotating about a fixed axis can be determined using the following procedure. Angular Motion. • Establish the positive sense of rotation about the axis of rotation and show it alongside each kinematic equation as it is applied.

• If a relation is known between any two of the four variables a, v, u,



and t, then a third variable can be obtained by using one of the following kinematic equations which relates all three variables. du dv v = a = a du = v dv dt dt If the body’s angular acceleration is constant, then the following equations can be used: v = v0 + act u = u0 + v0t + 12 act2 v2 = v20 + 2ac(u - u0)

• Once the solution is obtained, the sense of u, v, and a is

determined from the algebraic signs of their numerical quantities.

Motion of Point P. • In most cases the velocity of P and its two components of acceleration can be determined from the scalar equations v = vr at = ar an = v2r

• If the geometry of the problem is difficult to visualize, the following vector equations should be used: v = V * rP = V * r at = A * rP = A * r an = V * (V * rP) = -v2r

• Here rP is directed from any point on the axis of rotation to

point P, whereas r lies in the plane of motion of P. Either of these vectors, along with V and A, should be expressed in terms of its i, j, k components, and, if necessary, the cross products determined using a determinant expansion (see Eq. B–12).

16

328

Chapter 16

p l a n a r K i n e m at i C s

rigid Body

16.1

EXAMPlE

O

A cord is wrapped around a wheel in Fig. 16–6, which is initially at rest when u = 0. If a force is applied to the cord and gives it an acceleration a = (4t) m>s2, where t is in seconds, determine, as a function of time, (a) the angular velocity of the wheel, and (b) the angular position of line OP in radians.

0.2 m u P a

F

Fig. 16–6

16

of a

Solution part (a). The wheel is subjected to rotation about a fixed axis passing through point O. Thus, point P on the wheel has motion about a circular path, and the acceleration of this point has both tangential and normal components. The tangential component is (aP)t = (4t) m>s2, since the cord is wrapped around the wheel and moves tangent to it. Hence the angular acceleration of the wheel is (c +)

(aP)t = ar (4t) m>s2 = a(0.2 m) a = (20t) rad>s2b

Using this result, the wheel’s angular velocity v can now be determined from a = dv>dt, since this equation relates a, t, and v. Integrating, with the initial condition that v = 0 when t = 0, yields (c +)

dv = (20t) rad>s2 dt

a = L0

t

v

20t dt L0 v = 10t2 rad>sb

dv =

Ans.

part (b). Using this result, the angular position u of OP can be found from v = du>dt, since this equation relates u, v, and t. Integrating, with the initial condition u = 0 when t = 0, we have (c +)

du = v = (10t2) rad>s dt L0

u

du =

L0

t

10t2 dt

u = 3.33t3 rad

Ans.

NoTe: We cannot use the equation of constant angular acceleration,

since a is a function of time.

16.3

EXAMPlE

rotation aBout a fixed axis

329

16.2

The motor shown in the photo is used to turn a wheel and attached blower contained within the housing. The details are shown in Fig. 16–7a. If the pulley A connected to the motor begins to rotate from rest with a constant angular acceleration of aA = 2 rad>s2, determine the magnitudes of the velocity and acceleration of point P on the wheel, after the pulley has turned two revolutions. Assume the transmission belt does not slip on the pulley and wheel. Solution angular Motion. First we will convert the two revolutions to radians. Since there are 2p rad in one revolution, then uA = 2 rev a

2p rad b = 12.57 rad 1 rev

0.15 m A

aA  2 rad/s2

Since aA is constant, the angular velocity of pulley A is therefore

16

v2 = v20 + 2ac(u - u0)

(c +)

v2A = 0 + 2(2 rad>s2)(12.57 rad - 0) 0.4 m

vA = 7.090 rad>s The belt has the same speed and tangential component of acceleration as it passes over the pulley and wheel. Thus, v = vArA = vBrB;

B

P

7.090 rad>s (0.15 m) = vB(0.4 m)

(a)

vB = 2.659 rad>s at = aArA = aBrB;

2 rad>s2 (0.15 m) = aB(0.4 m) aB = 0.750 rad>s2

Motion of P. As shown on the kinematic diagram in Fig. 16–7b, we have vP = vBrB = 2.659 rad>s (0.4 m) = 1.06 m>s

aP

Ans.

(aP)t = aBrB = 0.750 rad>s2 (0.4 m) = 0.3 m>s2

(aP)n (aP)t

(aP)n = v2BrB = (2.659 rad>s)2(0.4 m) = 2.827 m>s2

vP

Thus

P

(b) 2 2

2 2

aP = 2(0.3 m>s ) + (2.827 m>s ) = 2.84 m>s

2

Ans.

Fig. 16–7

330

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

FundaMental prObleMs F16–1. When the gear rotates 20 revolutions, it achieves an angular velocity of v = 30 rad>s, starting from rest. Determine its constant angular acceleration and the time required.

F16–4. The bucket is hoisted by the rope that wraps around a drum wheel. If the angular displacement of the wheel is u = (0.5t3 + 15t) rad, where t is in seconds, determine the velocity and acceleration of the bucket when t = 3 s. 0.2 m

v

16

u

u

v

Prob. F16–1 F16–2. The flywheel rotates with an angular velocity of v = (0.005u2) rad>s, where u is in radians. Determine the angular acceleration when it has rotated 20 revolutions.

v

u

Prob. F16–4 F16–5. A wheel has an angular acceleration of a = (0.5 u) rad>s2, where u is in radians. Determine the magnitude of the velocity and acceleration of a point P located on its rim after the wheel has rotated 2 revolutions. The wheel has a radius of 0.2 m and starts at v0 = 2 rad>s. F16–6. For a short period of time, the motor turns gear A with a constant angular acceleration of aA = 4.5 rad>s2, starting from rest. Determine the velocity of the cylinder and the distance it travels in three seconds. The cord is wrapped around pulley D which is rigidly attached to gear B.

Prob. F16–2 F16–3. The flywheel rotates with an angular velocity of v = (4 u1>2) rad>s, where u is in radians. Determine the time it takes to achieve an angular velocity of v = 150 rad>s. When t = 0, u = 1 rad.

225 mm

75 mm B

aA  4.5 rad/s2 A

P

D

v u C

Prob. F16–3

Prob. F16–6

P¿

125 mm

16.3

331

rotation aBout a fixed axis

prObleMs 16–1. The angular velocity of the disk is defined by v = (5t2 + 2) rad>s, where t is in seconds. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = 0.5 s.

16–5. The disk is driven by a motor such that the angular position of the disk is defined by u = (20t + 4t2) rad, where t is in seconds. Determine the number of revolutions, the angular velocity, and angular acceleration of the disk when t = 90 s.

A

u 0.8 m

Prob. 16–5

Prob. 16–1

16–2. The angular acceleration of the disk is defined by a = 3t2 + 12 rad>s2, where t is in seconds. If the disk is originally rotating at v0 = 12 rad>s, determine the magnitude of the velocity and the n and t components of acceleration of point A on the disk when t = 2 s. 16–3. The disk is originally rotating at v0 = 12 rad >s. If  it is subjected to a constant angular acceleration of a = 20 rad >s2, determine the magnitudes of the velocity and the n and t components of acceleration of point A at the instant t = 2 s. *16–4. The disk is originally rotating at v0 = 12 rad>s. If it is subjected to a constant angular acceleration of a = 20 rad>s2, determine the magnitudes of the velocity and the n and t components of acceleration of point B when the disk undergoes 2 revolutions.

16–6. A wheel has an initial clockwise angular velocity of 10 rad>s and a constant angular acceleration of 3 rad>s2. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity of 15 rad>s. What time is required? 16–7. If gear A rotates with a constant angular acceleration of aA = 90 rad>s2, starting from rest, determine the time required for gear D to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear D to attain this angular velocity. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively. *16–8. If gear A rotates with an angular velocity of vA = (uA + 1) rad>s, where uA is the angular displacement of gear A, measured in radians, determine the angular acceleration of gear D when uA = 3 rad, starting from rest. Gears A, B, C, and D have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

D

v0  12 rad/s

A

B F

0.4 m 0.5 m

A

Probs. 16–2/3/4

B C

Probs. 16–7/8

16

332

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

16–9. At the instant vA = 5 rad>s, pulley A is given an angular acceleration a = (0.8u) rad>s2, where u is in radians. Determine the magnitude of acceleration of point B on pulley C when A rotates 3 revolutions. Pulley C has an inner hub which is fixed to its outer one and turns with it.

*16–12. The power of a bus engine is transmitted using the belt-and-pulley arrangement shown. If the engine turns pulley A at vA = (20t + 40) rad>s, where t is in seconds, determine the angular velocities of the generator pulley B and the air-conditioning pulley C when t = 3 s.

16–10. At the instant vA = 5 rad>s, pulley A is given a constant angular acceleration aA = 6 rad>s2. Determine the magnitude of acceleration of point B on pulley C when A rotates 2 revolutions. Pulley C has an inner hub which is fixed to its outer one and turns with it.

16–13. The power of a bus engine is transmitted using the belt-and-pulley arrangement shown. If the engine turns pulley A at vA = 60 rad>s, determine the angular velocities of the generator pulley B and the air-conditioning pulley C. The hub at D is rigidly connected to B and turns with it. 100 mm 25 mm

A

vA aA

vB D 75 mm

50 mm

B vC

16 40 mm

C B 60 mm

Probs. 16–9/10

16–11. The cord, which is wrapped around the disk, is given an acceleration of a = (10t) m>s2, where t is in seconds. Starting from rest, determine the angular displacement, angular velocity, and angular acceleration of the disk when t = 3 s.

a  (10t) m/s2

vA

C

Probs. 16–12/13 16–14. The disk starts from rest and is given an angular acceleration a = (2t 2) rad>s2, where t is in seconds. Determine the angular velocity of the disk and its angular displacement when t = 4 s. 16–15. The disk starts from rest and is given an angular acceleration a = (5t1> 2) rad>s2, where t is in seconds. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when t = 2 s. *16–16. The disk starts at v0 = 1 rad>s when u = 0, and is given an angular acceleration a = (0.3u) rad>s2, where u is in radians. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when u = 1 rev.

P

0.5 m 0.4 m

Prob. 16–11

50 mm

A

Probs. 16–14/15/16

16.3 16–17. A motor gives gear A an angular acceleration of aA = (2 + 0.006 u2) rad>s2, where u is in radians. If this gear is initially turning at vA = 15 rad>s, determine the angular velocity of gear B after A undergoes an angular displacement of 10 rev.

rotation aBout a fixed axis

333

*16–20. A motor gives gear A an angular acceleration of aA = (4t3) rad>s2, where t is in seconds. If this gear is initially turning at (vA )0 = 20 rad>s, determine the angular velocity of gear B when t = 2 s.

16–18. A motor gives gear A an angular acceleration of aA = (2t3) rad>s2, where t is in seconds. If this gear is initially turning at vA = 15 rad>s, determine the angular velocity of gear B when t = 3 s.

B

(vA)0  20 rad/s

A

175 mm 100 mm

0.05 m

aA vA

aB

B

A 0.15 m aA

16 Prob. 16–20

Probs. 16–17/18 16–19. Morse Industrial manufactures the speed reducer shown. If a motor drives the gear shaft S with an angular acceleration a = (4v - 3) rad>s2, where v is in rad>s, determine the angular velocity of shaft E at time t = 2 s after starting from an angular velocity 1 rad>s when t = 0. The radius of each gear is listed in the figure. Note that gears B and C are fixed connected to the same shaft.

16–21. The motor turns the disk with an angular velocity of v = (5t2 + 3t) rad>s, where t is in seconds. Determine the magnitudes of the velocity and the n and t components of acceleration of the point A on the disk when t = 3 s.

D B E C A

rA

20 mm

rB

80 mm

rC

30 mm

rD

120 mm 150 mm

S u

A

Prob. 16–19

Prob. 16–21

334

Chapter 16

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of a

16–22. If the motor turns gear A with an angular acceleration of aA = 2 rad>s2 when the angular velocity is vA = 20 rad>s, determine the angular acceleration and angular velocity of gear D.

rigid Body *16–24. The 50-mm-radius pulley A of the clothes dryer 2 rotates with an angular acceleration aA = (27u1>2 A ) rad>s , where uA is in radians. Determine its angular acceleration when t = 1 s, starting from rest. 16–25. If the 50-mm-radius motor pulley A of the clothes  dryer rotates with an angular acceleration of aA = (10 + 50t) rad>s2, where t is in seconds, determine its angular velocity when t = 3 s, starting from rest.

B

100 mm C 50 mm

D

16

vA 40 mm

A

100 mm 50 mm

Prob. 16–22

A

Probs. 16–24/25

16–23. If the motor turns gear A with an angular acceleration of aA = 3 rad>s2 when the angular velocity is vA = 60 rad>s, determine the angular acceleration and angular velocity of gear D.

B

16–26. The pinion gear A on the motor shaft is given a constant angular acceleration a = 3 rad>s2. If the gears A and B have the dimensions shown, determine the angular velocity and angular displacement of the output shaft C, when t = 2 s starting from rest. The shaft is fixed to B and turns with it.

100 mm B

C

C

50 mm

125 mm A

D

A

vA 40 mm

100 mm

Prob. 16–23

Prob. 16–26

35 mm

16.3 16–27. A stamp S, located on the revolving drum, is used to label canisters. If the canisters are centered 200 mm apart on the conveyor, determine the radius rA of the driving wheel A and the radius rB of the conveyor belt drum so that for each revolution of the stamp it marks the top of a canister. How many canisters are marked per minute if the drum at B is rotating at vB = 0.2 rad>s? Note that the driving belt is twisted as it passes between the wheels.

335

rotation aBout a fixed axis

16–29. For a short time a motor of the random-orbit sander drives the gear A with an angular velocity of vA = 40(t3 + 6t) rad>s, where t is in seconds. This gear is connected to gear B, which is fixed connected to the shaft CD. The end of this shaft is connected to the eccentric spindle EF and pad P, which causes the pad to orbit around shaft CD at a radius of 15 mm. Determine the magnitudes of the velocity and the tangential and normal components of acceleration of the spindle EF when t = 2 s after starting from rest.

40 mm

A rA

10 mm

B

S

VA A

C

200 mm

16

15 mm

D

rB

E

B F

rB

vB  0.2 rad/s

P

Prob. 16–27 Prob. 16–29 *16–28. At the instant shown, gear A is rotating with a constant angular velocity of vA = 6 rad>s. Determine the largest angular velocity of gear B and the maximum speed of point C.

16–30. Determine the distance the load W is lifted in t = 5 s using the hoist. The shaft of the motor M turns with an angular velocity v = 100(4 + t) rad>s, where t is in seconds. 300 mm

C

30 mm 225 mm

50 mm

A

M

D E C

100 mm vB

B

vA  6 rad/s

100 mm B

A

W 100 mm

100 mm

Prob. 16–28

40 mm

Prob. 16–30

336

Chapter 16

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of a

16–31. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If the angular displacement of A is uA = (5t3 + 10t2) rad, where t is in seconds, determine the angular velocity and angular acceleration of B when t = 3 s.

rigid Body 16–33. The rope of diameter d is wrapped around the tapered drum which has the dimensions shown. If the drum is rotating at a constant rate of v, determine the upward acceleration of the block. Neglect the small horizontal displacement of the block.

L r1 200 mm B

vA

r2

125 mm

vB

ω

d

A

16 Prob. 16–31 Prob. 16–33

*16–32. The driving belt is twisted so that pulley B rotates in the opposite direction to that of drive wheel A. If A has a constant angular acceleration of aA = 30 rad>s2, determine the tangential and normal components of acceleration of a  point located at the rim of B when t = 3 s, starting from rest.

16–34. A tape having a thickness s wraps around the wheel which is turning at a constant rate v. Assuming the unwrapped portion of tape remains horizontal, determine the acceleration of point P of the unwrapped tape when the radius of the wrapped tape is r. Hint: Since yP = vr, take the time derivative and note that dr>dt = v(s>2p).

s P aP r

200 mm B

vA

125 mm

vB A

v

Prob. 16–32

Prob. 16–34

16.3 16–35. If the shaft and plate rotates with a constant angular velocity of v = 14 rad>s, determine the velocity and acceleration of point C located on the corner of the plate at the instant shown. Express the result in Cartesian vector form.

337

rotation aBout a fixed axis

16–37. The rod assembly is supported by ball-and-socket joints at A and B. At the instant shown it is rotating about the y axis with an angular velocity v = 5 rad>s and has an angular acceleration a = 8 rad>s2. Determine the magnitudes of the velocity and acceleration of point C at this instant. Solve the problem using Cartesian vectors and Eqs. 16–9 and 16–13.

z A v

z

C

0.6 m

0.3 m

a 0.2 m C

D 0.4 m

O

0.4 m

0.3 m 0.3 m

x

B A

y

0.4 m

y

v

0.4 m

x

B

a

16

Prob. 16–37 Prob. 16–35

*16–36. At the instant shown, the shaft and plate rotates with an angular velocity of v = 14 rad>s and angular acceleration of a = 7 rad>s2. Determine the velocity and acceleration of point D located on the corner of the plate at this instant. Express the result in Cartesian vector form.

16–38. The mechanism for a car window winder is shown in the figure. Here the handle turns the small cog C, which rotates the spur gear S, thereby rotating the fixed-connected lever AB which raises track D in which the window rests. The window is free to slide on the track. If the handle is wound at 0.5 rad>s, determine the speed of points A and E and the speed vw of the window at the instant u = 30°.

z vw

A v

D

0.6 m

a

20 mm 0.2 m C

D 0.4 m

O

0.4 m B

Prob. 16–36

A

50 mm SB

0.3 m 0.3 m

x

0.5 rad/s

C

200 mm

y F

u

200 mm

Prob. 16–38

E

338

Chapter 16

p l a n a r K i n e m at i C s

of a

16.4

C A

b

u s a

B

16 The dumping bin on the truck rotates about a fixed axis passing through the pin at A. It is operated by the extension of the hydraulic cylinder BC. The angular position of the bin can be specified using the angular position coordinate u, and the position of point C on the bin is specified using the rectilinear position coordinate s. Since a and b are fixed lengths, then the two coordinates can be related by the cosine law, s = 2a2 + b2 - 2ab cos u. The time derivative of this equation relates the speed at which the hydraulic cylinder extends to the angular velocity of the bin.

rigid Body

Absolute Motion Analysis

A body subjected to general plane motion undergoes a simultaneous translation and rotation. If the body is represented by a thin slab, the slab translates in the plane of the slab and rotates about an axis perpendicular to this plane. The motion can be completely specified by knowing both the angular rotation of a line fixed in the body and the motion of a point on the body. One way to relate these motions is to use a rectilinear position coordinate s to locate the point along its path and an angular position coordinate u to specify the orientation of the line. The two coordinates are then related using the geometry of the problem. By direct application of the time-differential equations v = ds>dt, a = dv>dt, v = du>dt, and a = dv>dt, the motion of the point and the angular motion of the line can then be related. This procedure is similar to that used to solve dependent motion problems involving pulleys, Sec. 12.9. In some cases, this same procedure may be used to relate the motion of one body, undergoing either rotation about a fixed axis or translation, to that of a connected body undergoing general plane motion.

procedure for analysis The velocity and acceleration of a point P undergoing rectilinear motion can be related to the angular velocity and angular acceleration of a line contained within a body using the following procedure. Position Coordinate Equation.

• Locate point P on the body using a position coordinate s, which is

measured from a fixed origin and is directed along the straight-line path of motion of point P.

• Measure from a fixed reference line the angular position u of a line lying in the body.

• From the dimensions of the body, relate s to u, s = f(u), using geometry and/or trigonometry.

time Derivatives. • Take the first derivative of s = f(u) with respect to time to get a relation between v and v.

• Take the second time derivative to get a relation between a and a. • In each case the chain rule of calculus must be used when taking

the time derivatives of the position coordinate equation. See Appendix C.

16.4

EXAMPlE

339

aBsolute motion analysis

16.3 The end of rod R shown in Fig. 16–8 maintains contact with the cam by means of a spring. If the cam rotates about an axis passing through point O with an angular acceleration A and angular velocity V, determine the velocity and acceleration of the rod when the cam is in the arbitrary position u. V A

A r

r

u R

B

C x

O

16

Fig. 16–8

Solution position Coordinate equation. Coordinates u and x are chosen in order to relate the rotational motion of the line segment OA on the cam to the rectilinear translation of the rod. These coordinates are measured from the fixed point O and can be related to each other using trigonometry. Since OC = CB = r cos u, Fig. 16–8, then x = 2r cos u time derivatives.

Using the chain rule of calculus, we have

dx du = -2r(sin u) dt dt v = -2rv sin u

Ans.

dv dv du = -2r a b sin u - 2rv(cos u) dt dt dt 2 a = -2r(a sin u + v cos u)

Ans.

NoTe: The negative signs indicate that v and a are opposite to the

direction of positive x. This seems reasonable when you visualize the motion.

340

Chapter 16

EXAMPlE

p l a n a r K i n e m at i C s

of a

rigid Body

16.4 At a given instant, the cylinder of radius r, shown in Fig. 16–9, has an angular velocity V and angular acceleration A. Determine the velocity and acceleration of its center G if the cylinder rolls without slipping. V A

sG G¿ u

G r u A¿ A

B sG  ru

16

Fig. 16–9

Solution position Coordinate equation. The cylinder undergoes general plane motion since it simultaneously translates and rotates. By inspection, point G moves in a straight line to the left, from G to G, as the cylinder rolls, Fig. 16–9. Consequently its new position G will be specified by the horizontal position coordinate sG , which is measured from G to G. Also, as the cylinder rolls (without slipping), the arc length AB on the rim which was in contact with the ground from A to B, is equivalent to sG. Consequently, the motion requires the radial line GA to rotate u to the position GA. Since the arc AB = ru, then G travels a distance sG = ru time derivatives. Taking successive time derivatives of this equation, realizing that r is constant, v = du>dt, and a = dv>dt, gives the necessary relationships: sG = ru vG = rv

Ans.

aG = ra

Ans.

NoTe: Remember that these relationships are valid only if the cylinder

(disk, wheel, ball, etc.) rolls without slipping.

16.4

EXAMPlE

341

aBsolute motion analysis

16.5

The large window in Fig. 16–10 is opened using a hydraulic cylinder AB. If the cylinder extends at a constant rate of 0.5 m>s, determine the angular velocity and angular acceleration of the window at the instant u = 30. Solution position Coordinate equation. The angular motion of the window can be obtained using the coordinate u, whereas the extension or motion along the hydraulic cylinder is defined using a coordinate s, which measures its length from the fixed point A to the moving point B. These coordinates can be related using the law of cosines, namely, s2 = (2 m)2 + (1 m)2 - 2(2 m)(1 m) cos u

O 1m u

B 1m

2m

s

A

s2 = 5 - 4 cos u

(1)

When u = 30,

Fig. 16–10

s = 1.239 m time derivatives. Taking the time derivatives of Eq. 1, we have ds du = 0 - 4( -sin u) dt dt s(vs) = 2(sin u)v

2s

(2)

Since vs = 0.5 m>s, then at u = 30, (1.239 m)(0.5 m>s) = 2 sin 30v v = 0.6197 rad>s = 0.620 rad>s

Ans.

Taking the time derivative of Eq. 2 yields dvs ds du dv vs + s = 2(cos u) v + 2(sin u) dt dt dt dt 2 2 vs + sas = 2(cos u)v + 2(sin u)a Since as = dvs >dt = 0, then

(0.5 m>s)2 + 0 = 2 cos 30(0.6197 rad>s)2 + 2 sin 30a a = -0.415 rad>s2

Ans.

Because the result is negative, it indicates the window has an angular deceleration.

16

342

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

prObleMs 16–39. The end A of the bar is moving downward along the slotted guide with a constant velocity vA. Determine the angular velocity V and angular acceleration A of the bar as a function of its position y.

16–41. At the instant u = 50, the slotted guide is moving upward with an acceleration of 3 m>s2 and a velocity of 2 m>s. Determine the angular acceleration and angular velocity of link AB at this instant. Note: The upward motion of the guide is in the negative y direction.

B vA

300 mm v, a

A

y u

V, A U

16

A

y

r B v  2 m/s a  3 m/s2

Prob. 16–39 Prob. 16–41 *16–40. At the instant u = 60°, the slotted guide rod is moving to the left with an acceleration of 2 m>s2 and a velocity of 5 m>s. Determine the angular acceleration and angular velocity of link AB at this instant.

v  5 m/s a  2 m/s2

16–42. At the instant shown, u = 60°, and rod AB is  subjected to a deceleration of 16 m>s2 when the velocity is 10  m>s. Determine the angular velocity and angular acceleration of link CD at this instant.

x

v  10 m/s A

A

B

D u

u

u

a  16 m/s2 200 mm B

300 mm

300 mm

C

Prob. 16–40

Prob. 16–42

16–43. The crank AB is rotating with a constant angular velocity of 4 rad>s. Determine the angular velocity of the connecting rod CD at the instant u = 30.

343

aBsolute motion analysis

16.4

16–46. The circular cam rotates about the fixed point O with a constant angular velocity V. Determine the velocity v of the follower rod AB as a function of u.

C v R

u d

600 mm O

B

v

u

300 mm A

B

A

r

Prob. 16–46

4 rad/s

16–47. Determine the velocity of the rod R for any angle u of cam C as the cam rotates with a constant angular velocity V. The pin connection at O does not cause an interference with the motion of plate A on C.

D

Prob. 16–43

16

*16–44. Determine the velocity and acceleration of the follower rod CD as a function of u when the contact between the cam and follower is along the straight region AB on the face of the cam. The cam rotates with a constant counterclockwise angular velocity V.

V

r

C

v

A r O

u

D

C

u

R

B

A

O

x

Prob. 16–44

Prob. 16–47

16–45. Determine the velocity of rod R for any angle u of the cam C if the cam rotates with a constant angular velocity  V. The pin connection at O does not cause an interference with the motion of A on C.

*16–48. Determine the velocity and acceleration of the peg A which is confined between the vertical guide and the rotating slotted rod.

v

r1

C

A v a

r2 R

u

O

O

u

A x

Prob. 16–45

b

Prob. 16–48

344

Chapter 16

p l a n a r K i n e m at i C s

of a

16–49. Bar AB rotates uniformly about the fixed pin A with a constant angular velocity V. Determine the velocity and acceleration of block C, at the instant u = 60.

rigid Body 16–51. The pins at A and B are confined to move in the vertical and horizontal tracks. If the slotted arm is causing A to move downward at vA, determine the velocity of B at the instant shown.

B d L

v A

L

u

y

u

90

A

h

vA

B

C

16

x

L

Prob. 16–51

Prob. 16–49

16–50. The center of the cylinder is moving to the left with a constant velocity v0. Determine the angular velocity V and angular acceleration A of the bar. Neglect the thickness of the bar.

*16–52. The crank AB has a constant angular velocity V. Determine the velocity and acceleration of the slider at C as a function of u. Suggestion: Use the x coordinate to express the motion of C and the f coordinate for CB. x = 0 when f = 0°.

y B V

A u

l

r vO O

C

b v f

A x

Prob. 16–50

u

Prob. 16–52

x

16.4 16–53. If the wedge moves to the left with a constant velocity v, determine the angular velocity of the rod as a function of u.

aBsolute motion analysis

345

16–55. Arm AB has an angular velocity of V and an angular acceleration of A. If no slipping occurs between the disk D and the fixed curved surface, determine the angular velocity and angular acceleration of the disk.

v¿, a¿

D

Ar

L v

C

f

u

ω, a B

Prob. 16–53

R

16

Prob. 16–55

16–54. The crate is transported on a platform which rests on rollers, each having a radius r. If the rollers do not slip, determine their angular velocity if the platform moves forward with a velocity v.

*16–56. The bridge girder G of a bascule bridge is raised and lowered using the drive mechanism shown. If the hydraulic cylinder AB shortens at a constant rate of 0.15 m>s, determine the angular velocity of the bridge girder at the instant u = 60.

G

B v C

A

u 3m

5m r

v

Prob. 16–54

Prob. 16–56

346

Chapter 16

p l a n a r K i n e m at i C s

y¿

16.5

y A rB/A

x¿ Translating reference

rA B rB x

O

of a

Fixed reference

rigid Body

Relative-Motion Analysis: Velocity

The general plane motion of a rigid body can be described as a combina tion of translation and rotation. To view these “component” motions separately we will use a relative-motion analysis involving two sets of coordinate axes. The x, y coordinate system is fixed and measures the absolute position of two points A and B on the body, here represented as a bar, Fig. 16–11a. The origin of the x, y coordinate system will be attached to the selected “base point” A, which generally has a known motion. The axes of this coordinate system translate with respect to the fixed frame but do not rotate with the bar.

Position. The position vector rA in Fig. 16–11a specifies the location

(a)

of the “base point” A, and the relative-position vector rB>A locates point B with respect to point A. By vector addition, the position of B is then

Fig. 16–11

rB = rA + rB>A

Displacement. During an instant of time dt, points A and B

16

undergo displacements drA and drB as shown in Fig. 16–11b. If we consider the general plane motion by its component parts then the entire bar first translates by an amount drA so that A, the base point, moves to its final position and point B moves to B, Fig. 16–11c. The bar is then rotated about A by an amount du so that B undergoes a relative displacement drB>A and thus moves to its final position B. Due to the rotation about A, drB>A = rB>A du, and the displacement of B is drB = drA + drB>A due to rotation about A due to translation of A due to translation and rotation y¿

y¿ drA

A

A drA A

x¿

rB/A rB/A B

drB

Time t

B Time t  dt

General plane motion (b)

rB/A

du

B¿ dr B/A

drA B

A

B

drB Translation

(c)

Rotation

x¿

16.5

relative-motion analysis: veloCity

347

vB B C

V

A

vA

As slider block A moves horizontally to the left with a velocity vA , it causes crank CB to rotate counterclockwise, such that vB is directed tangent to its circular path, i.e., upward to the left. The connecting rod AB is subjected to general plane motion, and at the instant shown it has an angular velocity V.

Velocity. To determine the relation between the velocities of points A

and B, it is necessary to take the time derivative of the position equation, or simply divide the displacement equation by dt. This yields drB>A drB drA = + dt dt dt

The terms drB >dt = vB and drA >dt = vA are measured with respect to the fixed x, y axes and represent the absolute velocities of points A and B, respectively. Since the relative displacement is caused by a rotation, the # magnitude of the third term is drB>A >dt = rB>A du>dt = rB>Au = rB>Av, where v is the angular velocity of the body at the instant considered. We will denote this term as the relative velocity vB>A , since it represents the velocity of B with respect to A as measured by an observer fixed to the translating x, y axes. In other words, the bar appears to move as if it were rotating with an angular velocity V about the z axis passing through A. Consequently, vB>A has a magnitude of vB>A = vrB>A and a direction which is perpendicular to rB>A . We therefore have vB = vA + vB>A where vB = velocity of point B vA = velocity of the base point A vB>A = velocity of B with respect to A

(16–15)

16

348

Chapter 16

p l a n a r K i n e m at i C s

rigid Body

of a

Path of point A vA

A A vB

A

vB/A  vrB/A

vA

V





B

B

vA

B Path of point B

Rotation about the base point A

Translation

General plane motion

V rB/A

(e)

(d)

(f)

Fig. 16–11 (cont.) vB/A

vB

vA

16

(g)

B A

v

BC

45

C

vB = vA + V * rB>A

(a)

45 A

B

vA (b)

Fig. 16–12 V

A vA  0

Fig. 16–13

(16–16)

where vB

B

What the equation vB = vA + vB>A states is that the velocity of B, Fig. 16–11d, is determined by considering the entire bar to translate with a velocity of vA , Fig. 16–11e, and rotate about A with an angular velocity V, Fig. 16–11f. Vector addition of these two effects, applied to B, yields vB , as shown in Fig. 16–11g. Since the relative velocity vB>A represents the effect of circular motion, about A, this term can be expressed by the cross product vB>A = V * rB>A , Eq. 16–9. Hence, for application using Cartesian vector analysis, we can also write Eq. 16–15 as

vB

vB = velocity of B vA = velocity of the base point A V = angular velocity of the body rB>A = position vector directed from A to B The velocity equation 16–15 or 16–16 may be used in a practical manner to study the general plane motion of a rigid body which is either pin connected to or in contact with other moving bodies. When applying this equation, points A and B should generally be selected as points on the body which are pin-connected to other bodies, or as points in contact with adjacent bodies which have a known motion. For example, point A on link AB in Fig. 16–12a must move along a horizontal path, whereas point B moves on a circular path. The directions of vA and vB can therefore be established since they are always tangent to their paths of motion, Fig. 16–12b. In the case of the wheel in Fig. 16–13, which rolls without slipping, point A on the wheel can be selected at the ground. Here A (momentarily) has zero velocity since the ground does not move. Furthermore, the center of the wheel, B, moves along a horizontal path so that vB is horizontal.

16.5

relative-motion analysis: veloCity

349

procedure for analysis The relative velocity equation can be applied either by using Cartesian vector analysis, or by writing the x and y scalar component equations directly. For application, it is suggested that the following procedure be used. Vector Analysis Kinematic Diagram.

• Establish the directions of the fixed x, y coordinates and draw a

kinematic diagram of the body. Indicate on it the velocities vA , vB of points A and B, the angular velocity V, and the relativeposition vector rB>A .

• If the magnitudes of vA , vB , or V are unknown, the sense of direction of these vectors can be assumed.

Velocity Equation. • To apply vB = vA + V * rB>A , express the vectors in Cartesian vector form and substitute them into the equation. Evaluate the cross product and then equate the respective i and j components to obtain two scalar equations.

• If the solution yields a negative answer for an unknown magnitude, it indicates the sense of direction of the vector is opposite to that shown on the kinematic diagram.

Scalar Analysis Kinematic Diagram. • If the velocity equation is to be applied in scalar form, then the magnitude and direction of the relative velocity vB>A must be established. Draw a kinematic diagram such as shown in Fig. 16–11g, which shows the relative motion. Since the body is considered to be “pinned” momentarily at the base point A, the magnitude of vB>A is vB>A = vrB>A . The sense of direction of vB>A is always perpendicular to rB>A in accordance with the rotational motion V of the body.* Velocity Equation. • Write Eq. 16–15 in symbolic form, vB = vA + vB>A , and underneath each of the terms represent the vectors graphically by showing their magnitudes and directions. The scalar equations are determined from the x and y components of these vectors. *The notation vB = vA + vB>A(pin) may be helpful in recalling that A is “pinned.”

16

Chapter 16

p l a n a r K i n e m at i C s

rigid Body

16.6

EXAMPlE

The link shown in Fig. 16–14a is guided by two blocks at A and B, which move in the fixed slots. If the velocity of A is 2 m>s downward, determine the velocity of B at the instant u = 45.

vA  2 m/s A

0.2 m

Solution i (VECtor AnAlySiS) Kinematic diagram. Since points A and B are restricted to move along the fixed slots and vA is directed downward, then velocity vB must be directed horizontally to the right, Fig. 16–14b. This motion causes the link to rotate counterclockwise; that is, by the right-hand rule the angular velocity V is directed outward, perpendicular to the plane of motion.

B

u  45

0.1 m

C

velocity equation. Expressing each of the vectors in Fig. 16–14b in terms of their i, j, k components and applying Eq. 16–16 to A, the base point, and B, we have vB = vA + V * rB>A vBi = -2j + [vk * (0.2 sin 45i - 0.2 cos 45j)] vBi = -2j + 0.2v sin 45j + 0.2v cos 45i

(a)

16

y A

x

rB/A V

vA  2 m/s 45

Equating the i and j components gives vB = 0.2v cos 45 0 = -2 + 0.2v sin 45 Thus, v = 14.1 rad>sd

vB

B

(b)

rB/A V

vB/A 45

B

Relative motion (c)

Fig. 16–14

vB = 2 m>s S

Ans.

Solution ii (SCAlAr AnAlySiS) The kinematic diagram of the relative “circular motion” which produces vB>A is shown in Fig. 16–14c. Here vB>A = v(0.2 m). Thus, vB = vA + vB/A

A

45

of a

+ ) (S

c

vB 2 m>s v(0.2 m) d = c d + c d S T 45 vB = 0 + v(0.2) cos 45 d

350

(+ c ) 0 = -2 + v(0.2) sin 45 The solution produces the above results. It should be emphasized that these results are valid only at the instant u = 45. A recalculation for u = 44 yields vB = 2.07 m>s and v = 14.4 rad>s; whereas when u = 46, vB = 1.93 m>s and v = 13.9 rad>s, etc. NoTe: Since vA and v are known, the velocity of any other point on the link can be determined. As an exercise, see if you can apply Eq. 16–16 to points A and C or to points B and C and show that when u = 45, vC = 3.16 m>s, directed at an angle of 18.4 up from the horizontal.

16.5

351

relative-motion analysis: veloCity

16.7

EXAMPlE

The cylinder shown in Fig. 16–15a rolls without slipping on the surface of a conveyor belt which is moving at 0.6 m>s. Determine the velocity of point A. The cylinder has a clockwise angular velocity v = 15 rad>s at the instant shown.

y v  15 rad/s x

0.15 m A

O

Solution i (VECtor AnAlySiS) Kinematic diagram. Since no slipping occurs, point B on the cylinder has the same velocity as the conveyor, Fig. 16–15b. Also, the angular velocity of the cylinder is known, so we can apply the velocity equation to B, the base point, and A to determine vA .

vC  0.6 m/s

B

(a)

velocity equation vA = vB + V * rA>B (vA)xi + (vA)y j = 0.6i + ( -15k) * ( -0.5i + 0.5j)

v  15 rad/s

vA u

A

(vA)xi + (vA)y j = 0.6i + 2.25j + 2.25i

16

rA/B B

so that (vA)x = 0.6 + 2.25 = 2.85 m>s

(1)

(vA)y = 2.25 m>s

(2)

(b)

vB  0.6 m/s

Thus, vA = 2(2.85)2 + (2.25)2 = 3.63 m>s 2.25 b = 38.3 2.85

d

u = tan-1 a

Ans. Ans.

Solution ii (SCAlAr AnAlySiS) As an alternative procedure, the scalar components of vA = vB + vA>B can be obtained directly. From the kinematic diagram showing the relative “circular” motion which produces vA>B , Fig. 16–15c, we have 0.15 m vA>B = vrA>B = (15 rad>s) a b = 3.18 m>s cos 45 Thus, vA = vB + vA>B (vA)x (v ) 0.6 m>s 3.18 m>s d + c A yd = c d + c d S S c 45 d

c

Equating the x and y components gives the same results as before, namely, + ) (S (vA)x = 0.6 + 3.18 cos 45 = 2.85 m>s (+ c ) (vA)y = 0 + 3.18 sin 45 = 2.25 m>s

vA/B A

45 45 rA/B

B

v  15 rad/s

0.15 m

Relative motion (c)

Fig. 16–15

352

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

16.8

EXAMPlE

The collar C in Fig. 16–16a is moving downward with a velocity of 2 m>s. Determine the angular velocity of CB at this instant. C A

vC  2 m/s

0.2 m

Solution i (VECtor AnAlySiS) Kinematic diagram. The downward motion of C causes B to move to the right along a curved path. Also, CB and AB rotate counterclockwise. velocity equation. link CB (general plane motion): See Fig. 16–16b.

B 0.2 m

vB = vC + VCB * rB>C

(a)

vBi = -2j + vCBk * (0.2i - 0.2j) vBi = -2j + 0.2vCB j + 0.2vCBi

y C VCB

B

vCB = 10 rad>s d

VCB

45

rB/C

vB/A

(2) Ans.

vB = 2 m>s S

vB

(b) C

0 = -2 + 0.2vCB

rB/C

vC  2 m/s

(1)

vB = 0.2vCB

x

Solution ii (SCAlAr AnAlySiS) The scalar component equations of vB = vC + vB>C can be obtained directly. The kinematic diagram in Fig. 16–16c shows the relative “circular” motion which produces vB>C . We have vB = vC + vB>C

45 B Relative motion

c

(c)

vB 2 m>s v 1 0.222 m 2 d = c d + c CB d S T 45 d

16

Resolving these vectors in the x and y directions yields + ) (S

A VAB

(+ c )

0.2 m

vB = 0 + vCB 1 0.222 cos 45 2

0 = -2 + vCB 1 0.222 sin 45 2

which is the same as Eqs. 1 and 2. vB  2 m/s

B (d)

Fig. 16–16

NoTe: Since link AB rotates about a fixed axis and vB is known,

Fig. 16–16d, its angular velocity is found from vB = vABrAB or 2 m>s = vAB (0.2 m), vAB = 10 rad>s.

16.5

353

relative-motion analysis: veloCity

preliMinary prObleM P16–1. Set up the relative velocity equation between points A and B.

3m A

2m

B

30 3m

3 rad/s A

6 rad/s (d) 60

2m

B

16

(a) B

4 rad/s

v A

4 rad/s B 30

0.5 m

0.5 m

3m

No slipping

A

(e)

No slipping (b)

B A

4m 3m

A 30

2 rad/s

1m

B

6 rad/s

45 4m

4m

(f)

(c)

Prob. P16–1

354

Chapter 16

p l a n a r K i n e m at i C s

of a

rigid Body

FundaMental prObleMs F16–7. If roller A moves to the right with a constant velocity of vA = 3 m>s, determine the angular velocity of the link and the velocity of roller B at the instant u = 30.

F16–10. If crank OA rotates with an angular velocity of v = 12 rad>s, determine the velocity of piston B and the angular velocity of rod AB at the instant shown. A 0.6 m

B 0.3 m 12 rad/s

1.5 m

30

O

u  30

A

Prob. F16–10

vA  3 m/s

16

B

Prob. F16–7 F16–8. The wheel rolls without slipping with an angular velocity of v = 10 rad>s. Determine the magnitude of the velocity of point B at the instant shown.

F16–11. If rod AB slides along the horizontal slot with a velocity of 18 m>s, determine the angular velocity of link BC at the instant shown. 0.15 m O 30

C

0.75 m 18 m/s

0.6 m

A

B

v

Prob. F16–11

B

F16–12. End A of the link has a velocity of vA = 3 m>s. Determine the velocity of the peg at B at this instant. The peg is constrained to move along the slot.

A

Prob. F16–8 F16–9. Determine the angular velocity of the spool. The cable wraps around the inner core, and the spool does not slip on the platform P. 0.6 m

0.6 m/s

A

2m

1.2 m/s

B O

vA  3 m/s

45

0.3 m

B A

P 30

Prob. F16–9

Prob. F16–12

16.5

355

relative-motion analysis: veloCity

prObleMs 16–57. The wheel is rotating with an angular velocity v = 8 rad>s. Determine the velocity of the collar A at the instant u = 30 and f = 60. Also, sketch the location of bar AB when u = 0, 30, and 60 to show its general plane motion.

*16–60. At the instant shown, the truck is traveling to the right at 8 m>s. If the spool does not slip at B, determine its angular velocity so that its mass center G appears to an observer on the ground to remain stationary.

A f

16–59. At the instant shown, the truck is traveling to the right at 3 m>s, while the pipe is rolling counterclockwise at angular v = 8 rad>s without slipping at B. Determine the velocity of the pipe’s center G.

60 500 mm

30

u

150 mm

B

v

V

8 rad/s

16

G 1.5 m

O B

Prob. 16–57

Probs. 16–59/60

16–58. The slider block C moves at 8 m>s down the inclined groove. Determine the angular velocities of links AB and BC, at the instant shown.

2m A

16–61. The link AB has an angular velocity of 3 rad>s. Determine the velocity of block C and the angular velocity of link BC at the instant u = 45. Also, sketch the position of link BC when u = 60, 45, and 30° to show its general plane motion.

B

45 2m

vA B  3 rad/s

B

1.5 m

0.5 m C

u  45 vC  8 m/s

Prob. 16–58

A

Prob. 16–61

C

356

Chapter 16

p l a n a r K i n e m at i C s

of a

16–62. If the gear rotates with an angular velocity of v = 10 rad>s and the gear rack moves at yC = 5 m>s, determine the velocity of the slider block A at the instant shown.

rigid Body *16–64. The cylinder B rolls on the fixed cylinder A without slipping. If the connected bar CD is rotating with an angular velocity of vCD = 5 rad>s, determine the angular velocity of cylinder B.

A 0.3 m D

0.1 m C 0.5 m

v

10 rad/s B

60

16

vC

5 rad/s

VCD

A

Prob. 16–64

75 mm

5 m/s C

Prob. 16–62

16–63. Knowing that angular velocity of link AB is vAB = 4 rad>s, determine the velocity of the collar at C and the angular velocity of link CB at the instant shown. Link CB is horizontal at this instant.

16–65. The angular velocity of link AB is vAB = 5 rad>s. Determine the velocity of block C and the angular velocity of link BC at the instant u = 45° and f = 30°. Also, sketch the position of link CB when u = 45°, 60°, and 75° to show its general plane motion.

A

350 mm

C

vA B  5 rad/s

B

3m u B

45 vAB  4 rad/s

500 mm

f

2m

60 A

Prob. 16–63

C

Prob. 16–65

16.5 16–66. The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the velocity of the slider block C at the instant u = 60, if link AB is rotating at 4 rad>s.

357

relative-motion analysis: veloCity

16–69. If the slider block C is moving at vC = 3 m>s, determine the angular velocity of BC and the crank AB at the instant shown.

16–67. Determine the velocity of the slider block at C at the instant u = 45, if link AB is rotating at 4 rad>s. B 0.5 m 125 mm C 45

A

1m

60

45

B

C vC

vAB

3 m/s

16

Prob. 16–69

4 rad/s 300 mm u

A

16–70. Determine the velocity of the center O of the spool when the cable is pulled to the right with a velocity of v. The spool rolls without slipping.

Probs. 16–66/67

*16–68. Rod AB is rotating with an angular velocity of vAB = 60 rad>s. Determine the velocity of the slider C at the instant u = 60° and f = 45°. Also, sketch the position of  bar BC when u = 30°, 60° and 90° to show its general plane motion.

16–71. Determine the velocity of point A on the outer rim of the spool at the instant shown when the cable is pulled to the right with a velocity of v. The spool rolls without slipping.

300 mm B

600 mm f

u

A

A vAB  60 rad/s

O

r

R C

Prob. 16–68

Probs. 16–70/71

v

358

Chapter 16

p l a n a r K i n e m at i C s

of a

*16–72. If the flywheel is rotating with an angular velocity of vA = 6 rad>s, determine the angular velocity of rod BC at the instant shown.

rigid Body 16–74. If the slider block A is moving downward at vA = 4 m>s, determine the velocities of blocks B and C at the instant shown. 16–75. If the slider block A is moving downward at vA = 4 m>s, determine the velocity of point E at the instant shown.

1.5 m A

B O

6 rad/s

vA

0.3 m

250 mm

3 5

D

400 mm

C

vA  4 m/s

300 mm

B

30 60 0.6 m

D

E

300 mm

4

A

C

Probs. 16–74/75

16 Prob. 16–72

16–73. The epicyclic gear train consists of the sun gear A which is in mesh with the planet gear B. This gear has an inner hub C which is fixed to B and in mesh with the fixed ring gear R. If the connecting link DE pinned to B and C is rotating at vDE = 18 rad>s about the pin at E, determine the angular velocities of the planet and sun gears.

*16–76. The planetary gear system is used in an automatic transmission for an automobile. By locking or releasing certain gears, it has the advantage of operating the car at different speeds. Consider the case where the ring gear R is held fixed, vR = 0, and the sun gear S is rotating at vS = 5 rad>s. Determine the angular velocity of each of the planet gears P and shaft A.

40 mm vR P

R

100 mm 600 mm

A E

vS

B C D

S

A

80 mm

200 mm vDE  18 rad/s 300 mm

R

Prob. 16–73

40 mm

Prob. 16–76

16.5 16–77. The mechanism is used on a machine for the manufacturing of a wire product. Because of the rotational motion of link AB and the sliding of block F, the segmental gear lever DE undergoes general plane motion. If AB is rotating at vAB = 5 rad>s, determine the velocity of point E at the instant shown.

16–79. If the ring gear A rotates clockwise with an angular velocity of vA = 30 rad>s, while link BC rotates clockwise with an angular velocity of vBC = 15 rad>s, determine the angular velocity of gear D.

A E

50 mm 45

359

relative-motion analysis: veloCity

vA 30 rad/s

20 mm

D

F C

D

C

vBC  15 rad/s 250 mm

300 mm

20 mm 50 mm

200 mm

B 45

B

A

16

Prob. 16–79

45

vAB  5 rad/s

Prob. 16–77 16–78. The similar links AB and CD rotate about the fixed  pins at A and C. If AB has an angular velocity vAB = 8 rad>s, determine the angular velocity of BDP and the velocity of point P.

B

300 mm

300 mm

*16–80. The mechanism shown is used in a riveting machine. It consists of a driving piston A, three links, and a riveter which is attached to the slider block D. Determine the velocity of D at the instant shown, when the piston at A is traveling at vA = 20 m>s.

D 300 mm

300 mm

C

A vAB  8 rad/s

C

60

60

45 150 mm

700 mm

300 mm vA  20 m/s

45 200 mm

D B

45

P

Prob. 16–78

A

30

60

Prob. 16–80

360

Chapter 16

p l a n a r K i n e m at i C s

of a

16.6

16

rigid Body

Instantaneous Center of Zero Velocity

The velocity of any point B located on a rigid body can be obtained in a very direct way by choosing the base point A to be a point that has zero velocity at the instant considered. In this case, vA = 0, and therefore the velocity equation, vB = vA + V * rB>A , becomes vB = V * rB>A . For a body having general plane motion, point A so chosen is called the instantaneous center of zero velocity (IC), and it lies on the instantaneous axis of zero velocity. This axis is always perpendicular to the plane of motion, and the intersection of the axis with this plane defines the location of the IC. Since point A coincides with the IC, then vB = V * rB>IC and so point B moves momentarily about the IC in a circular path; in other words, the body appears to rotate about the instantaneous axis. The magnitude of vB is simply vB = vrB>IC , where v is the angular velocity of the body. Due to the circular motion, the direction of vB must always be perpendicular to rB>IC . For example, the IC for the bicycle wheel in Fig. 16–17 is at the contact point with the ground. There the spokes are somewhat visible, whereas at the top of the wheel they become blurred. If one imagines that the wheel is momentarily pinned at this point, the velocities of various points can be found using v = vr. Here the radial distances shown in the photo, Fig. 16–17, must be determined from the geometry of the wheel.

IC

Fig. 16–17

16.6

361

Instantaneous Center of Zero VeloCIty

Centrode

IC

vIC  0 A

vA V

rA/IC

rA/IC

vIC  0 V

A

IC

vA rB/IC B

Location of IC knowing vA and V

vB

(a)

Location of IC knowing the directions of vA and vB

A

16

rA/IC

(b)

Fig. 16–18

vA

V

IC

d

Location of the IC. To locate the IC we can use the fact that the

velocity of a point on the body is always perpendicular to the relativeposition vector directed from the IC to the point. Several possibilities exist:

• The velocity vA of a point A on the body and the angular velocity V

rB/IC vB

B

of the body are known, Fig. 16–18a. In this case, the IC is located along the line drawn perpendicular to vA at A, such that the distance from A to the IC is rA>IC = vA >v. Note that the IC lies up and to the right of A since vA must cause a clockwise angular velocity V about the IC.

(c)

IC rA/IC

• The lines of action of two nonparallel velocities vA and vB are known, Fig. 16–18b. Construct at points A and B line segments that are perpendicular to vA and vB . Extending these perpendiculars to their point of intersection as shown locates the IC at the instant considered.

• The magnitude and direction of two parallel velocities vA and vB are known. Here the location of the IC is determined by proportional triangles. Examples are shown in Fig. 16–18c and d. In both cases rA>IC = vA >v and rB>IC = vB >v. If d is a known distance between points A and B, then in Fig. 16–18c, rA>IC + rB>IC = d and in Fig. 16–18d, rB>IC - rA>IC = d.

rB/IC

A B

vA d vB

V Location of IC knowing vA and vB (d)

362

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

As the board slides downward to the left it is subjected to general plane motion. Since the directions of the velocities of its ends A and B are known, the IC is located as shown. At this instant the board will momentarily rotate about this point. Draw the board in several other positions and establish the IC for each case.

IC

B vB

vA A

IC

16 rA/IC rB/IC

rC/IC

B V A

vB  v rB/IC C

vA  v rA/IC

Fig. 16–19

Realize that the point chosen as the instantaneous center of zero velocity for the body can only be used at the instant considered since the body changes its position from one instant to the next. The locus of points which define the location of the IC during the body’s motion is called a centrode, Fig. 16–18a, and so each point on the centrode acts as the IC for the body only for an instant. Although the IC may be conveniently used to determine the velocity of any point in a body, it generally does not have zero acceleration and therefore it should not be used for finding the accelerations of points in a body.

vC  v rC/IC

Procedure for Analysis The velocity of a point on a body which is subjected to general plane motion can be determined with reference to its instantaneous center of zero velocity provided the location of the IC is first established using one of the three methods described above. • As shown on the kinematic diagram in Fig. 16–19, the body is imagined as “extended and pinned” at the IC so that, at the instant considered, it rotates about this pin with its angular velocity V.

• The magnitude of velocity for each of the arbitrary points A, B, and C on the body can be determined by using the equation v = vr, where r is the radial distance from the IC to each point.

• The line of action of each velocity vector v is perpendicular to its

associated radial line r, and the velocity has a sense of direction which tends to move the point in a manner consistent with the angular rotation V of the radial line, Fig. 16–19.

16.6

EXAMPLE

363

Instantaneous Center of Zero VeloCIty

16.9

Show how to determine the location of the instantaneous center of zero velocity for (a) member BC shown in Fig. 16–20a; and (b) the link CB shown in Fig. 16–20c. IC rB/IC

VBC

B

rC/IC

u A

B v

C

u vB

C

(a)

vC

(b)

SoLution Part (a). As shown in Fig. 16–20a, point B moves in a circular path such that vB is perpendicular to AB. Therefore, it acts at an angle u from the horizontal as shown in Fig. 16–20b. The motion of point B causes the piston to move forward horizontally with a velocity vC . When lines are drawn perpendicular to vB and vC , Fig. 16–20b, they intersect at the IC. Part (b). Points B and C follow circular paths of motion since links  AB and DC are each subjected to rotation about a fixed axis, Fig.  16–20c. Since the velocity is always tangent to the path, at the instant considered, vC on rod DC and vB on rod AB are both directed vertically downward, along the axis of link CB, Fig. 16–20d. Radial lines  drawn perpendicular to these two velocities form parallel lines which intersect at “infinity;” i.e., rC>IC S  and rB>IC S . Thus, vCB = (vC >rC>IC) S 0. As a result, link CB momentarily translates. An instant later, however, CB will move to a tilted position, causing the IC to move to some finite location. vC rC/IC

C

VCB

IC rB/IC

B vB

(d)

Fig. 16–20

16

b D C

VDC

A

B

(c)

364

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

16.10

EXAMPLE

Block D shown in Fig. 16–21a moves with a speed of 3 m>s. Determine the angular velocities of links BD and AB, at the instant shown.

B 0.4 m

0.4 m 90

45

A

45

3 m/s D

(a)

IC

16

rB/IC vB rD/IC

B VBD 45 0.4 m

D

vD  3 m/s

SoLution As D moves to the right, it causes AB to rotate clockwise about point A. Hence, vB is directed perpendicular to AB. The instantaneous center of zero velocity for BD is located at the intersection of the line segments drawn perpendicular to vB and vD , Fig. 16–21b. From the geometry, rB>IC = 0.4 tan 45 m = 0.4 m rD>IC =

(b)

0.4 m = 0.5657 m cos 45

Since the magnitude of vD is known, the angular velocity of link BD is vBD =

vD rD>IC

=

3 m>s = 5.30 rad>sd 0.5657 m

Ans.

The velocity of B is therefore B

45 vB  2.12 m/s

0.4 m

vB = vBD(rB>IC) = 5.30 rad>s (0.4 m) = 2.12 m>s d 45 From Fig. 16–21c, the angular velocity of AB is

VAB A

vAB = (c)

Fig. 16–21

2.12 m>s vB = = 5.30 rad>sb rB>A 0.4 m

Ans.

Note: Try to solve this problem by applying vD = vB + vD>B to member BD.

16.6

365

Instantaneous Center of Zero VeloCIty

16.11

EXAMPLE

The cylinder shown in Fig. 16–22a rolls without slipping between the two moving plates E and D. Determine the angular velocity of the cylinder and the velocity of its center C.

E

A

0.125 m vD = 0.4 m/s

vE = 0.25 m/s

C

B

D

(a)

16

SoLution Since no slipping occurs, the contact points A and B on the cylinder have the same velocities as the plates E and D, respectively. Furthermore, the velocities vA and vB are parallel, so that by the proportionality of right triangles the IC is located at a point on line AB, Fig. 16–22b. Assuming this point to be a distance x from B, we have vB = vx;

0.4 m>s = vx

vA = v(0.25 m - x);

0.25 m>s = v(0.25 m - x)

Hence, the angular velocity of the cylinder is Ans.

The velocity of point C is therefore vC = vrC>IC = 2.60 rad>s (0.1538 m - 0.125 m) = 0.0750 m>s d

rC/IC rC C

0.125 m

Fig. 16–22

0.1 = 0.1538 m 0.65

0.4 m>s vB = = 2.60 rad>sb x 0.1538 m

IC

(b)

0.4(0.25 - x) = 0.25x

v =

vA  0.25 m/s

vB  0.4 m/s B

Dividing one equation into the other eliminates v and yields

x =

A V

Ans.

0.25 m x

366

Chapter 16

EXAMPLE

p l a n a r K I n e m at I C s

of a

rIgId Body

16.12 The crankshaft AB turns with a clockwise angular velocity of 10 rad>s, Fig. 16–23a. Determine the velocity of the piston at the instant shown. C

0.75 m

13.6 VBC

B 45

0.25 m

VAB  10 rad/s A

16 (a)

SoLution The crankshaft rotates about a fixed axis, and so the velocity of point B is vB = (10 rad>s) (0.25 m) = 2.50 m>s 45 d

vC IC

45.0

76.4

C

0.75 m 58.6 2.50 m/s B (b)

Fig. 16–23

Since the directions of the velocities of B and C are known, then the location of the IC for the connecting rod BC is at the intersection of the lines extended from these points, perpendicular to vB and vC , Fig. 16–23b. The magnitudes of rB>IC and rC>IC can be obtained from the geometry of the triangle and the law of sines, i.e., rB>IC 0.75 m = sin 45 sin 76.4 rB>IC = 1.031 m rC>IC 0.75 m = sin 45 sin 58.6 rC>IC = 0.9056 m The rotational sense of VBC must be the same as the rotation caused by vB about the IC, which is counterclockwise. Therefore, 2.5 m>s vB vBC = = = 2.425 rad>s rB>IC 1.031 m Using this result, the velocity of the piston is vC = vBCrC>IC = (2.425 rad>s)(0.9056 m) = 2.20 m>s Ans.

16.6

367

Instantaneous Center of Zero VeloCIty

PreliminAry Problem P16–2. Establish the location of the instantaneous center of zero velocity for finding the velocity of point B.

4 m/s

C 1m 8 rad/s

B

B

45

1m A

2m

No slipping

3 m/s

(a)

(d)

2m

16

A 3

4 rad/s

5 4

3 rad/s

0.5 m B

A

2m

B

0.5 m

(b)

0.5 m

No slipping

No slipping (e)

A

0.5 m

1.5 m 4 rad/s 30

0.5 m

B

0.3 m

30 45

B (c)

A

2m (f)

Prob. P16–2

6 rad/s

368

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

FundAmentAl Problems F16–13. Determine the angular velocity of the rod and the velocity of point C at the instant shown.

F16–16. If cable AB is unwound with a speed of 3 m>s, and the gear rack C has a speed of 1.5 m>s, determine the angular velocity of the gear and the velocity of its center O.

vA  6 m/s 3 m/s

A

0.3 m 2.5 m

4m

C

A

B

0.2 m

O C

1.5 m/s

2.5 m

Prob. F16–16 B

16

F16–17. Determine the angular velocity of link BC and the velocity of the piston C at the instant shown.

Prob. F16–13

F16–14. Determine the angular velocity of link BC and velocity of the piston C at the instant shown.

A

0.8 m

0.2 m

vAB  12 rad/s B

30

B C

A

C v  6 rad/s

0.6 m

1.2 m

Prob. F16–17

Prob. F16–14

F16–15. If the center O of the wheel is moving with a speed of vO = 6 m>s, determine the velocity of point A on the wheel. The gear rack B is fixed.

F16–18. Determine the angular velocity of links BC and CD at the instant shown.

D 0.2 m

0.4 m 0.6 m A

0.2 m

0.3 m O

6 m/s B

Prob. F16–15

A

B 30 vAB  10 rad/s

Prob. F16–18

C

16.6

369

Instantaneous Center of Zero VeloCIty

Problems 16–81. In each case show graphically how to locate the instantaneous center of zero velocity of link AB. Assume the geometry is known.

16–83. The square plate is confined within the slots at A and B. When u = 30, point A is moving at vA = 8 m>s. Determine the velocity of point C at this instant. *16–84. The square plate is confined within the slots at A and B. When u = 30, point A is moving at vA = 8 m>s. Determine the velocity of point D at this instant. D

B

0.3 m

v 0.3 m

A

A

B

(a) v

v

C A

(b)

B

C B u

(c)

A

Prob. 16–81

vA

30

8 m/s

Probs. 16–83/84

16–82. If crank AB is rotating with an angular velocity of vAB = 6 rad>s, determine the velocity of the center O of the gear at the instant shown.

16–85. The shaper mechanism is designed to give a slow cutting stroke and a quick return to a blade attached to the slider at C. Determine the angular velocity of the link CB at the instant shown, if the link AB is rotating at 4 rad>s.

125 mm C 45

B

0.6 m

B

C

0.4 m

O vAB

A

6 rad/s

0.1 m

vAB  4 rad/s

300 mm

0.1 m 60

60 A

Prob. 16–82

Prob. 16–85

16

370

Chapter 16

p l a n a r K I n e m at I C s

of a

16–86. At the instant shown, the disk is rotating at v = 4 rad>s. Determine the velocities of points A, B, and C.

rIgId Body 16–89. If rod CD is rotating with an angular velocity vCD = 4 rad>s, determine the angular velocities of rods AB and CB at the instant shown.

D

A C

v

O

A

30

4 rad/s

0.4 m

1m

B

0.15 m

B 0.5 m

E

16

C

vC D  4 rad/s

Prob. 16–86

D

Prob. 16–89 16–87. Member AB is rotating at vAB = 6 rad>s. Determine the velocity of point D and the angular velocity of members BPD and CD.

*16–88. Member AB is rotating at vAB = 6 rad>s. Determine the velocity of point P, and the angular velocity of member BPD.

B

200 mm

200 mm

D 200 mm

200 mm 60 A

16–90. If bar AB has an angular velocity vAB = 6 rad>s, determine the velocity of the slider block C at the instant shown.

vAB  6 rad/s

60

250 mm

200 mm C

vA B  6 rad/s

B

A

u  45

500 mm 30 C

P

Probs. 16–87/88

Prob. 16–90

16.6 16–91. The mechanism used in a marine engine consists of a single crank AB and two connecting rods BC and BD. Determine the velocity of the piston at C the instant the crank is in the position shown and has an angular velocity of 5 rad>s.

371

Instantaneous Center of Zero VeloCIty

16–94. The pinion gear A rolls on the fixed gear rack B with an angular velocity v = 8 rad>s. Determine the velocity of the gear rack C.

C

*16–92. The mechanism used in a marine engine consists of a single crank AB and two connecting rods BC and BD. Determine the velocity of the piston at D the instant the crank is in the position shown and has an angular velocity of 5 rad>s.

A v

150 mm

B

Prob. 16–94 D 45

60 C

30

0.4 m

45 0.4 m

16–95. The cylinder B rolls on the fixed cylinder A without slipping. If connected bar CD is rotating with an angular velocity vCD = 5 rad>s, determine the angular velocity of 16 cylinder B. Point C is a fixed point.

B 0.2 m 5 rad/s A

Probs. 16–91/92

0.3 m

0.1 m D C

16–93. Show that if the rim of the wheel and its hub maintain contact with the three tracks as the wheel rolls, it is necessary that slipping occurs at the hub A if no slipping occurs at B. Under these conditions, what is the speed at A if the wheel has angular velocity V?

v

A

vCD  5 rad/s

B

Prob. 16–95 *16–96. Determine the angular velocity of the doubletooth gear and the velocity of point C on the gear.

r2 r1

vA A

A

4 m/s 0.3 m

0.15 m

C B

Prob. 16–93

vB B

Prob. 16–96

6 m/s

372

Chapter 16

p l a n a r K I n e m at I C s

of a

16–97. If the hub gear H and ring gear R have angular velocities vH = 5 rad>s and vR = 20 rad>s, respectively, determine the angular velocity vS of the spur gear S and the angular velocity of its attached arm OA.

rIgId Body *16–100. Cylinder A rolls on the fixed cylinder B without slipping. If bar CD is rotating with an angular velocity of vCD = 3 rad>s, determine the angular velocity of A.

16–98. If the hub gear H has an angular velocity vH = 5 rad>s, determine the angular velocity of the ring gear R so that the arm OA attached to the spur gear S remains stationary (vOA = 0). What is the angular velocity of the spur gear?

C 200 mm

vR 50 mm

250 mm

A

H

vCD S 200 mm

16

D B

vS

O O

150 mm

A

vH

Prob. 16–100

R

Probs. 16–97/98 16–99. The crankshaft AB rotates at vAB = 50 rad>s about the fixed axis through point A, and the disk at C is held fixed in its support at E. Determine the angular velocity of rod CD at the instant shown.

16–101. The planet gear A is pin connected to the end of the link BC. If the link rotates about the fixed point B at 4 rad>s, determine the angular velocity of the ring gear R. The sun gear D is fixed from rotating. 16–102. Solve Prob. 16–101 if the sun gear D is rotating clockwise at vD = 5 rad>s while link BC rotates counterclockwise at vBC = 4 rad>s.

E

C

75 mm 75 mm

40 mm

F

D

300 mm

R D vBC  4 rad/s B 150 mm

60

A C 75 mm

vAB  50 rad/s A

B 100 mm

Prob. 16–99

Probs. 16–101/102

vR

16.7

16.7

373

relatIVe-motIon analysIs: aCCeleratIon Path of point A

Relative-Motion Analysis: Acceleration

aA

A

An equation that relates the accelerations of two points on a bar (rigid body) subjected to general plane motion may be determined by differentiating vB = vA + vB>A with respect to time. This yields dvB>A dvB dvA = + dt dt dt

aB

VA

B

Path of point B

General plane motion (a)

aA



The terms dvB >dt = aB and dvA >dt = aA are measured with respect to a set of fixed x, y axes and represent the absolute accelerations of points B and A. The last term represents the acceleration of B with respect to A as measured by an observer fixed to translating x, y axes which have their origin at the base point A. In Sec. 16.5 it was shown that to this observer point B appears to move along a circular arc that has a radius of curvature rB>A . Consequently, aB>A can be expressed in terms of its tangential and normal components; i.e., aB>A = (aB>A)t + (aB>A)n , where (aB>A)t = arB>A and (aB>A)n = v2rB>A . Hence, the relative-acceleration equation can be written in the form

A

16

aA

aB = aA + (aB>A)t + (aB>A)n

(16–17)

B Translation

where

(b)

aB = acceleration of point B aA = acceleration of point A (aB>A)t = tangential acceleration component of B with respect to A. The magnitude is (aB>A)t = arB>A , and the direction is perpendicular to rB>A . (aB>A)n = normal acceleration component of B with respect to A. The magnitude is (aB>A)n = v2rB>A , and the direction is always from B toward A . The terms in Eq. 16–17 are represented graphically in Fig. 16–24. Here it is seen that at a given instant the acceleration of B, Fig. 16–24a, is determined by considering the bar to translate with an acceleration aA , Fig. 16–24b, and simultaneously rotate about the base point A with an instantaneous angular velocity V and angular acceleration A, Fig. 16–24c. Vector addition of these two effects, applied to B, yields aB , as shown in Fig. 16–24d. It should be noted from Fig. 16–24a that since points A and B move along curved paths, the accelerations of these points will have both tangential and normal components. (Recall that the acceleration of a point is tangent to the path only when the path is rectilinear or when it is an inflection point on a curve.)

A



V A

aB/A

rB/A (aB/A)n

(aB/A)t B Rotation about the base point A (c) (aB/A)n (aB/A)t aB aA (d)

Fig. 16–24

374

Chapter 16

Path of B

A

p l a n a r K I n e m at I C s

C

a

aB = aA + A * rB>A - v2rB>A

(16–18)

where

(a) B (aB)n

aB aA A V

(aB)t

aB

rB>A

A

16

rIgId Body

Since the relative-acceleration components represent the effect of circular motion observed from translating axes having their origin at the base point A, these terms can be expressed as (aB>A)t = A * rB>A and (aB>A)n = -v2rB>A , Eq. 16–14. Hence, Eq. 16–17 becomes

B

v

of a

B

aB

aC

C (b)

Fig. 16–25

V

= = = = =

acceleration of point B acceleration of the base point A angular acceleration of the body angular velocity of the body position vector directed from A to B

If Eq. 16–17 or 16–18 is applied in a practical manner to study the accelerated motion of a rigid body which is pin connected to two other bodies, it should be realized that points which are coincident at the pin move with the same acceleration, since the path of motion over which they travel is the same. For example, point B lying on either rod BA or BC of the crank mechanism shown in Fig. 16–25a has the same acceleration, since the rods are pin connected at B. Here the motion of B is along a circular path, so that aB can be expressed in terms of its tangential and normal components. At the other end of rod BC point C moves along a straight-lined path, which is defined by the piston. Hence, aC is horizontal, Fig. 16–25b. Finally, consider a disk that rolls without slipping as shown in Fig. 16–26a. As a result, vA = 0 and so from the kinematic diagram in Fig. 16–26b, the velocity of the mass center G is

A

vG = vA + V * rG>A = 0 + (-vk) * (rj)

G

So that

r

vG = vr

A (a) V G

vG

rG/A A (b)

Fig. 16–26

(16–19)

This same result can also be determined using the IC method where point A is the IC. Since G moves along a straight line, its acceleration in this case can be determined from the time derivative of its velocity. dvG dv = r dt dt aG = ar

(16–20)

These two important results were also obtained in Example 16–4. They apply as well to any circular object, such as a ball, gear, wheel, etc., that rolls without slipping.

16.7

relatIVe-motIon analysIs: aCCeleratIon

375

Procedure for Analysis B

The relative acceleration equation can be applied between any two points A and B on a body either by using a Cartesian vector analysis, or by writing the x and y scalar component equations directly. Velocity Analysis. • Determine the angular velocity V of the body by using a velocity analysis as discussed in Sec. 16.5 or 16.6. Also, determine the velocities vA and vB of points A and B if these points move along curved paths. Vector Analysis Kinematic Diagram. • Establish the directions of the fixed x, y coordinates and draw the kinematic diagram of the body. Indicate on it aA , aB , V, A, and rB>A . • If points A and B move along curved paths, then their accelerations should be indicated in terms of their tangential and normal components, i.e., aA = (aA)t + (aA)n and aB = (aB)t + (aB)n . Acceleration Equation. • To apply aB = aA + A * rB>A - v2rB>A, express the vectors in Cartesian vector form and substitute them into the equation. Evaluate the cross product and then equate the respective i and j components to obtain two scalar equations.

• If the solution yields a negative answer for an unknown magnitude, it indicates that the sense of direction of the vector is opposite to that shown on the kinematic diagram. Scalar Analysis Kinematic Diagram. • If the acceleration equation is applied in scalar form, then the magnitudes and directions of the relative-acceleration components (aB>A)t and (aB>A)n must be established. To do this draw a kinematic diagram such as shown in Fig. 16–24c. Since the body is considered to be momentarily “pinned” at the base point A, the magnitudes of these components are (aB>A)t = arB>A and (aB>A)n = v2rB>A . Their sense of direction is established from the diagram such that (aB>A)t acts perpendicular to rB>A , in accordance with the rotational motion A of the body, and (aB>A)n is directed from B toward A.* Acceleration Equation. • Represent the vectors in aB = aA + (aB>A)t + (aB>A)n graphically by showing their magnitudes and directions underneath each term. The scalar equations are determined from the x and y components of these vectors.

*The notation aB = aA + (aB>A(pin))t + (aB>A(pin))n may be helpful in recalling that A is assumed to be pinned.

V, A aB

(aA)t

A

(aA)n

C

The mechanism for a window is shown. Here CA rotates about a fixed axis through C, and AB undergoes general plane motion. Since point A moves along a curved path it has two components of acceleration, whereas point B moves along a straight track and the direction of its acceleration is specified.

16

376

Chapter 16

p l a n a r K I n e m at I C s

of a

16.13

EXAMPLE

The rod AB shown in Fig. 16–27a is confined to move along the inclined planes at A and B. If point A has an acceleration of 3 m>s2 and a velocity of 2 m>s, both directed down the plane at the instant the rod is horizontal, determine the angular acceleration of the rod at this instant.

10 m A

B

vA  2 m/s aA  3 m/s2 45

SoLution i (VEctor AnALySiS) We will apply the acceleration equation to points A and B on the rod. To do so it is first necessary to determine the angular velocity of the rod. Show that it is v = 0.283 rad>sd using either the velocity equation or the method of instantaneous centers.

45 (a)

y aB

x A

A 45

rIgId Body

rB/A

B

Kinematic diagram. Since points A and B both move along straight-line paths, they have no components of acceleration normal to the paths. There are two unknowns in Fig. 16–27b, namely, aB and a.

45

v  0.283 rad/s

16 aA  3 m/s2

Acceleration equation. (b)

aB = aA + A * rB>A - v2rB>A aB cos 45i + aB sin 45j = 3 cos 45i - 3 sin 45j + (ak) * (10i) - (0.283)2(10i) Carrying out the cross product and equating the i and j components yields

(aB/A)t  a rB/A

v  0.283 rad/s

B

aB sin 45 = -3 sin 45 + a(10)

(2)

Solving, we have aB = 1.87 m>s2 45

rB/A

a = 0.344 rad>s2 d

Ans.

(c)

Fig. 16–27

SoLution ii (ScALAr AnALySiS) From the kinematic diagram, showing the relative-acceleration components (aB>A)t and (aB>A)n , Fig. 16–27c, we have aB = aA + (aB>A)t + (aB>A)n c

aB 3 m>s2 a(10 m) (0.283 rad>s)2(10 m) d = c d + c d + c d d d 45 45 c

d

A

(1)

d

A

10 m (aB/A)n  v2 rB/A

aB cos 45 = 3 cos 45 - (0.283)2(10)

Equating the x and y components yields Eqs. 1 and 2, and the solution proceeds as before.

16.7

EXAMPLE

relatIVe-motIon analysIs: aCCeleratIon

377

16.14

The disk rolls without slipping and has the angular motion shown in Fig. 16–28a. Determine the acceleration of point A at this instant.

v  6 rad/s a  4 rad/s2 G 0.5 m

SoLution i (VEctor AnALySiS) Kinematic diagram. Since no slipping occurs, applying Eq. 16–20,

A

aG = ar = (4 rad>s2)(0.5 m) = 2 m>s2

(a)

Acceleration equation. We will apply the acceleration equation to points G and A, Fig. 16–28b,

v  6 rad/s a  4 rad/s2 2 m/s2 G

aA = aG + A : rA>G - v2rA>G aA = -2i + (4k) : (-0.5j) - (6)2(-0.5j) = {18j} m>s

rA/G (aA)y

2

(aA)x

A (b)

SoLution ii (ScALAr AnALySiS) Using the result for aG = 2 m>s2 determined above, and from the kinematic diagram, showing the relative motion aA>G, Fig. 16–28c, we have

v  6 rad/s a  4 rad/s2

(aA/G)y

aA = aG + (aA>G)x + (aA>G)y c

(aA)x (aA)y 2 m>s2 (4 rad>s2)(0.5 m) (6 rad>s)2(0.5 m) d = c d + c d d + c d + c S S d c c

+ S

(aA)x = -2 + 2 = 0

+c

(aA)y = 18 m>s2

Therefore, aA = 2(0)2 + (18 m>s2)2 = 18 m>s2

G

Ans.

Note: The fact that aA = 18 m>s2 indicates that the instantaneous

center of zero velocity, point A, is not a point of zero acceleration.

rA/G

A (c)

Fig. 16–28

(aA/G)x

16

378

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

16.15

EXAMPLE

The spool shown in Fig. 16–29a unravels from the cord, such that at the instant shown it has an angular velocity of 3 rad>s and an angular acceleration of 4 rad>s2. Determine the acceleration of point B. B

G

A

0.2 m

0.3 m v  3 rad/s a  4 rad/s2

SoLution i (VEctor AnALySiS) The spool “appears” to be rolling downward without slipping at point A. Therefore, we can use the results of Eq. 16–20 to determine the acceleration of point G, i.e., aG = ar = (4 rad>s2)(0.2 m) = 0.8 m>s2 We will apply the acceleration equation to points G and B. Kinematic diagram. Point B moves along a curved path having an unknown radius of curvature.* Its acceleration will be represented by its unknown x and y components as shown in Fig. 16–29b.

(a)

Acceleration equation.

16

aB = aG + A * rB>G - v2rB>G

y

(aB)xi + (aB)y j = -0.8j + ( -4k) * (0.3j) - (3)2(0.3j) x

(aB)y (aB)x

Equating the i and j terms, the component equations are (aB)x = 4(0.3) = 1.2 m>s2 S (aB)y = -0.8 - 2.7 = -3.5 m>s2 = 3.5 m>s2 T

rB/G

aG  0.8 m/s2

aB = 2(1.2)2 + (3.5)2 = 3.7 m>s2 u = tan-1 a

(b)

(aB/G)t  arB/G

G v  3 rad/s a  4 rad/s2 (c)

Fig. 16–29

3.5 b = 71.1 d 1.2

Ans. Ans.

SoLution ii (ScALAr AnALySiS) This problem may be solved by writing the scalar component equations directly. The kinematic diagram in Fig. 16–29c shows the relativeacceleration components (aB>G)t and (aB>G)n . Thus, aB = aG + (aB>G)t + (aB>G)n

(aB/G)n  v2rB/G rB/G  0.3 m

(2)

The magnitude and direction of aB are therefore

v  3 rad/s a  4 rad/s2

B

(1)

c

(aB)x (a ) d + c B yd S c

0.8 m>s2 4 rad>s2 (0.3 m) (3 rad>s)2(0.3 m) d + c d + c d S T T The x and y components yield Eqs. 1 and 2 above. *Realize that the path’s radius of curvature r is not equal to the radius of the spool since the spool is not rotating about point G. Furthermore, r is not defined as the distance = c

from A (IC) to B, since the location of the IC depends only on the velocity of a point and not the geometry of its path.

16.7

EXAMPLE

379

relatIVe-motIon analysIs: aCCeleratIon

16.16

The collar C in Fig. 16–30a moves downward with an acceleration of 1 m>s2. At the instant shown, it has a speed of 2 m>s which gives links CB and AB an angular velocity vAB = vCB = 10 rad>s. (See Example 16.8.) Determine the angular accelerations of CB and AB at this instant.

aC  1 m/s2 vC  2 m/s

C A

vAB 10 rad/s 0.2 m

vCB  10 rad/s B

16

0.2 m (a)

SoLution (VEctor AnALySiS) Kinematic diagram. The kinematic diagrams of both links AB and CB are shown in Fig. 16–30b. To solve, we will apply the appropriate kinematic equation to each link. Acceleration equation. Link AB (rotation about a fixed axis): aB = AAB * rB - v2ABrB aB = (aABk) * (-0.2j) - (10)2(-0.2j) aB = 0.2aABi + 20j

C

aC  1 m/s

A ACB rB/C vCB  10 rad/s

vAB  10 rad/s AAB rB

B

Thus, 0.2aAB = 0.2aCB - 20 20 = -1 + 0.2aCB + 20 aCB = 5 rad>s2 d aAB = -95 rad>s2 = 95 rad>s2 b

x

2

Note that aB has n and t components since it moves along a circular path. Link BC (general plane motion): Using the result for aB and applying Eq. 16–18, we have aB = aC + ACB * rB>C - v2CBrB>C 0.2aABi + 20j = -1j + (aCBk) * (0.2i - 0.2j) - (10)2(0.2i - 0.2j) 0.2aABi + 20j = -1j + 0.2aCB j + 0.2aCBi - 20i + 20j

Solving,

y

Ans. Ans.

0.2 m (b)

Fig. 16–30

0.2 m

380

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

16.17

EXAMPLE

The crankshaft AB turns with a clockwise angular acceleration of 20 rad>s2, Fig. 16–31a. Determine the acceleration of the piston at the instant AB is in the position shown. At this instant vAB = 10 rad>s and vBC = 2.43 rad>s. (See Example 16.12.) Please refer to the Companion Website for the animation: Kinematics of a Collar

C

Acceleration equation. Cartesian vector form

13.6

0.75 m

vBC  2.43 rad/s

16

B

45

vAB  10 rad/s aAB  20 rad/s2 A

0.25 m

SoLution (VEctor AnALySiS) Kinematic diagram. The kinematic diagrams for both AB and BC are shown in Fig. 16–31b. Here aC is vertical since C moves along a straight-line path. Expressing each of the position vectors in

rB = 5 -0.25 sin 45i + 0.25 cos 45j 6 m = 5 -0.177i + 0.177j 6 m

rC>B = 5 0.75 sin 13.6i + 0.75 cos 13.6j 6 m = 5 0.177i + 0.729j 6 m

crankshaft AB (rotation about a fixed axis): aB = AAB * rB - v2ABrB

= ( -20k) * (-0.177i + 0.177j) - (10)2(-0.177i + 0.177j) = 5 21.21i - 14.14j 6 m>s2

(a) y

connecting rod BC (general plane motion): Using the result for aB and noting that aC is in the vertical direction, we have aC C

aC = aB + ABC * rC>B - v2BCrC>B aCj = 21.21i - 14.14j + (aBCk) * (0.177i + 0.729j) - (2.43)2(0.177i + 0.729j) aCj = 21.21i - 14.14j + 0.177aBC j - 0.729aBCi - 1.04i - 4.30j

rC/B 13.6 aBC vBC  2.43 rad/s

0.75 cos 13.6 m B

45

0.25 cos 45 m r B

0 = 20.17 - 0.729aBC aC = 0.177aBC - 18.45 Solving yields aBC = 27.7 rad>s2 d

vAB  10 rad/s aAB  20 rad/s2 A x

aC = -13.5 m>s2

Ans.

Note: Since the piston is moving upward, the negative sign for aC (b)

Fig. 16–31

indicates that the piston is decelerating, i.e., aC = 5 -13.5j 6 m>s2. This causes the speed of the piston to decrease until AB becomes vertical, at which time the piston is momentarily at rest.

16.7

381

relatIVe-motIon analysIs: aCCeleratIon

PreliminAry Problem P16–3. Set up the relative acceleration equation between points A and B. The angular velocity is given. 2

6 m/s v  3 rad/s

A B

2m

B

3m

45

2m

60 (d)

v  2.12 rad/s

A

16

3 m/s

0.5 m

2

2 m/s

v  1.15 rad/s

(a) B v  4 rad/s

B 45

a  2 rad/s

2m

A 4 rad/s 2 8 rad/s

30 2m

2

A (e)

No slipping (b)

4m B

A

B v  4 rad/s

v0

2

1m

a  2 rad/s

6 rad/s

2m

0.5 m

3 rad/s 2 2 rad/s

(c)

A

(f)

Prob. P16–3

382

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

FundAmentAl Problems F16–19. At the instant shown, end A of the rod has the velocity and acceleration shown. Determine the angular acceleration of the rod and acceleration of end B of the rod. aA  5 m/s2 vA  6 m/s

F16–22. At the instant shown, cable AB has a velocity of 3 m>s and acceleration of 1.5 m>s2, while the gear rack has a velocity of 1.5 m>s and acceleration of 0.75 m>s2. Determine the angular acceleration of the gear at this instant. aB  1.5 m/s2 vB  3 m/s

A 0.3 m

B

0.2 m O

5m

4m

A

aC  0.75 m/s2 vC  1.5 m/s

C

Prob. F16–22

B

16 Prob. F16–19 F16–20. The gear rolls on the fixed rack with an angular velocity of v = 12 rad>s and angular acceleration of a = 6 rad>s2. Determine the acceleration of point A. A

a  6 rad/s2 v  12 rad/s

F16–23. At the instant shown, the wheel rotates with an angular velocity of v = 12 rad>s and an angular acceleration of a = 6 rad>s2. Determine the angular acceleration of link BC at the instant shown. D 0.3 m B

0.3 m O

0.3 m 45

C

1.2 m

a  6 rad/s2 v  12 rad/s

Prob. F16–23 Prob. F16–20 F16–21. The gear rolls on the fixed rack B. At the instant shown, the center O of the gear moves with a velocity of vO = 6 m>s and acceleration of aO = 3 m>s2. Determine the angular acceleration of the gear and acceleration of point A at this instant.

F16–24. At the instant shown, wheel A rotates with an angular velocity of v = 6 rad>s and an angular acceleration of a = 3 rad>s2. Determine the angular acceleration of link BC and the acceleration of piston C. 0.8 m 30

B 0.6 m A

O

0.3 m a  3 m/s2 O vO  6 m/s B

Prob. F16–21

0.2 m

C

A

v  6 rad/s a  3 rad/s2

Prob. F16–24

16.7

383

relatIVe-motIon analysIs: aCCeleratIon

Problems 16–103. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the angular acceleration of pulley B at the instant shown. *16–104. Pulley A rotates with the angular velocity and angular acceleration shown. Determine the acceleration of block E at the instant shown.

16–106. At a given instant the roller A on the bar has the velocity and acceleration shown. Determine the velocity and acceleration of the roller B, and the bar’s angular velocity and angular acceleration at this instant.

50 mm 4 m/s 6 m/s2

A 30

40 rad/s 5 rad/s2

vA aA

0.6 m

16 30

50 mm

B

B 125 mm E

Prob. 16–106

Probs. 16–103/104

16–107. Bar AB has the angular motions shown. Determine the velocity and acceleration of the slider block C at this instant.

16–105. Member AB has the angular motions shown. Determine the velocity and acceleration of the slider block C at this instant.

B

vAB  4 rad/s

B

0.5 m

aAB  6 rad/s2

A

45

1m

2m C

5

3

A

4 rad/s 5 rad/s2

60 C

0.5 m

4

Prob. 16–105

Prob. 16–107

384

Chapter 16

p l a n a r K I n e m at I C s

of a

*16–108. The flywheel rotates with angular velocity v = 2 rad>s and angular acceleration a = 6 rad>s2. Determine the angular acceleration of links AB and BC at this instant.

rIgId Body 16–110. The reel of rope has the angular motion shown. Determine the velocity and acceleration of point A at the instant shown. 16–111. The reel of rope has the angular motion shown. Determine the velocity and acceleration of point B at the instant shown.

A v a

2 rad/s 6 rad/s2

3

0.5 m B

4

0.3 m 0.4 m a  8 rad/s2 v  3 rad/s

C

16

C

B

Prob. 16–108

100 mm A

Probs. 16–110/111

16–109. A cord is wrapped around the inner spool of the gear. If it is pulled with a constant velocity v, determine the velocities and accelerations of points A and B. The gear rolls on the fixed gear rack.

*16–112. The disk has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point B.

v  3 rad/s a  8 rad/s2

B 2r A

r

G

0.5 m

45

C

v 45 A

Prob. 16–109

Prob. 16–112

B

16.7 16–113. The disk has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point C.

relatIVe-motIon analysIs: aCCeleratIon

385

16–115. Determine the angular acceleration of link CD if link AB has the angular velocity and angular acceleration shown.

D 0.5 m

0.5 m

v  3 rad/s a  8 rad/s2 C 0.5 m

45

45

C 1m

aAB  6 rad/s2 vAB  3 rad/s

B

A

A

16 B

Prob. 16–113

1m

Prob. 16–115

16–114. Member AB has the angular motions shown. Determine the angular velocity and angular acceleration of members CB and DC.

*16–116. At a given instant the slider block A is moving to the right with the motion shown. Determine the angular acceleration of link AB and the acceleration of point B at this instant.

vA  4 m/s aA  6 m/s2 A

30 B D

200 mm vAB  2 rad/s

C

2m

450 mm

60

100 mm

B

aAB  4 rad/s2

Prob. 16–114

2m A

Prob. 16–116

386

Chapter 16

p l a n a r K I n e m at I C s

of a

16–117. The slider block has the motion shown. Determine the angular velocity and angular acceleration of the wheel at this instant.

rIgId Body 16–119. If member AB has the angular motion shown, determine the angular velocity and angular acceleration of member CD at the instant shown. 300 mm A

A

B

150 mm

vAB  3 rad/s aAB  8 rad/s2

C 500 mm

400 mm

B

C

vB  4 m/s aB  2 m/s2

16

u  60

200 mm

Prob. 16–117

D

Prob. 16–119 16–118. The disk rolls without slipping such that it has an angular acceleration of a = 4 rad>s2 and angular velocity of v = 2 rad>s at the instant shown. Determine the acceleration of points A and B on the link and the link’s angular acceleration at this instant. Assume point A lies on the periphery of the disk, 150 mm from C.

*16–120. If member AB has the angular motion shown, determine the velocity and acceleration of point C at the instant shown. 300 mm A

B vAB  3 rad/s aAB  8 rad/s2

A 500 mm

v  2 rad/s a  4 rad/s2 500 mm

C

C

u  60

150 mm 200 mm

B D

400 mm

Prob. 16–118

Prob. 16–120

16.7 16–121. The wheel rolls without slipping such that at the instant shown it has an angular velocity V and angular acceleration A. Determine the velocity and acceleration of point B on the rod at this instant.

A

O

2a

387

relatIVe-motIon analysIs: aCCeleratIon

16–123. The collar is moving downward with the motion shown. Determine the angular velocity and angular acceleration of the gear at the instant shown as it rolls along the fixed gear rack.

A

v  2 m/s a  3 m/s2 v, a

500 mm

60

a

B

O

B

150 mm 200 mm

16 Prob. 16–121

Prob. 16–123

16–122. A single pulley having both an inner and outer rim is pin connected to the block at A. As cord CF unwinds from the inner rim of the pulley with the motion shown, cord DE unwinds from the outer rim. Determine the angular acceleration of the pulley and the acceleration of the block at the instant shown.

*16–124. The tied crank and gear mechanism gives rocking motion to crank AC, necessary for the operation of a printing press. If link DE has the angular motion shown, determine the respective angular velocities of gear F and crank AC at this instant, and the angular acceleration of crank AC.

C F

100 mm

50 mm

vDE  4 rad/s

75 mm 100 mm

B

D 25 mm

50 mm F

C

A

E D

G

E 150 mm

v F  2 m/s a F  3 m/s 2

Prob. 16–122

Prob. 16–124

aDE  20 rad/s2 30

A

388

Chapter 16

p l a n a r K I n e m at I C s

of a

16–125. The center O of the gear and the gear rack P move with the velocities and accelerations shown. Determine the angular acceleration of the gear and the acceleration of point B located at the rim of the gear at the instant shown.

rIgId Body 16–127. Determine the angular acceleration of link AB if link CD has the angular velocity and angular deceleration shown. 0.6 m C

0.6 m B

150 mm 3 m/s 6 m/s2

vO aO vP aP

B

O

2 m/s 3 m/s2

0.3 m

16

4 rad/s2

vCD

2 rad/s

D

A

P

A

aCD

Prob. 16–125

Prob. 16–127

16–126. The hoop is cast on the rough surface such that it has angular velocity v = 4 rad>s and an angular acceleration a = 5 rad>s2. Also, its center has a velocity vO = 5 m>s and a deceleration aO = 2 m>s2. Determine the acceleration of point A at this instant.

*16–128. The mechanism produces intermittent motion of link AB. If the sprocket S is turning with an angular acceleration aS = 2 rad>s2 and has an angular velocity vS = 6 rad>s at the instant shown, determine the angular velocity and angular acceleration of link AB at this instant. The sprocket S is mounted on a shaft which is separate from a collinear shaft attached to AB at A. The pin at C is attached to one of the chain links such that it moves vertically downward.

200 mm B

A A

v

4 rad/s a

aO

2 m/s2

5 rad/s2

aS  2 rad/s2

O 0.3 m

vS  6 rad/s

vO

45

C

S 175 mm

5 m/s D

B

Prob. 16–126

15 150 mm

30

50 mm

Prob. 16–128

16.8

16.8

389

relatIVe-motIon analysIs usIng rotatIng axes

Relative-Motion Analysis Using Rotating Axes

In the previous sections the relative-motion analysis for velocity and acceleration was described using a translating coordinate system. This type of analysis is useful for determining the motion of points on the same rigid body, or the motion of points located on several pin-connected bodies. In some problems, however, rigid bodies (mechanisms) are constructed such that sliding will occur at their connections. The kinematic analysis for such cases is best performed if the motion is analyzed using a coordinate system which both translates and rotates. Furthermore, this frame of reference is useful for analyzing the motions of two points on a mechanism which are not located in the same body and for specifying the kinematics of particle motion when the particle moves along a rotating path. In the following analysis two equations will be developed which relate the velocity and acceleration of two points, one of which is the origin of a moving frame of reference subjected to both a translation and a rotation in the plane.*

Position.

Consider the two points A and B shown in Fig. 16–32a. Their location is specified by the position vectors rA and rB , which are measured with respect to the fixed X, Y, Z coordinate system. As shown in the figure, the “base point” A represents the origin of the x, y, z coordinate system, which is assumed to be both translating and rotating with respect to the X, Y, Z system. The position of B with respect to A is specified by the relative-position vector rB>A . The components of this vector may be expressed either in terms of unit vectors along the X, Y axes, i.e., I and J, or by unit vectors along the x, y axes, i.e., i and j. For the development which follows, rB>A will be measured with respect to the moving x, y frame of reference. Thus, if B has coordinates (xB , yB), Fig. 16–32a, then rB>A = xBi + yBj

Y  y

xB

B

(16–21)

At the instant considered, point A has a velocity vA and an acceleration aA , while the angular# velocity and angular acceleration of the x, y axes are  (omega) and  = d >dt, respectively. *The more general, three-dimensional motion of the points is developed in Sec. 20.4.

yB

.

 x

rB/A rB A rA X (a)

Fig. 16–32

Using vector addition, the three position vectors in Fig. 16–32a are related by the equation rB = rA + rB>A

16

390

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

Velocity.

The velocity of point B is determined by taking the time derivative of Eq. 16–21, which yields vB = vA +

drB>A dt

(16–22)

The last term in this equation is evaluated as follows: drB>A dt

=

d (x i + yB j) dt B

=

dyB dj dxB di i + xB + j + yB dt dt dt dt

= a

y dj j¿

(16–23)

The two terms in the first set of parentheses represent the components of velocity of point B as measured by an observer attached to the moving x, y, z coordinate system. These terms will be denoted by vector (vB>A)xyz . In the second set of parentheses the instantaneous time rate of change of the unit vectors i and j is measured by an observer located in the fixed X, Y, Z coordinate system. These changes, di and dj, are due only to the rotation du of the x, y, z axes, causing i to become i = i + di and j to become j = j + dj, Fig. 16–32b. As shown, the magnitudes of both di and dj equal 1 du, since i = i = j = j = 1. The direction of di is defined by +j, since di is tangent to the path described by the arrowhead of i in the limit as t S dt. Likewise, dj acts in the -i direction, Fig. 16–32b. Hence,

16

du

dyB dj dxB di i + jb + axB + yB b dt dt dt dt

du x

j

i¿

j1

di du = (j) = j dt dt

di

i i1 

dj du = (-i) = - i dt dt

Viewing the axes in three dimensions, Fig. 16–32c, and noting that  = k, we can express the above derivatives in terms of the cross product as

(b) z

di =  * i dt

 x k y

i

dj =  * j dt

(16–24)

Substituting these results into Eq. 16–23 and using the distributive property of the vector cross product, we obtain

j (c)

Fig. 16–32 (cont.)

drB>A dt

= (vB>A)xyz +  * (xBi + yB j) = (vB>A)xyz +  * rB>A

(16–25)

16.8

relatIVe-motIon analysIs usIng rotatIng axes

391

Hence, Eq. 16–22 becomes vB = vA +  * rB>A + (vB>A)xyz

(16–26)

where vB = velocity of B, measured from the X, Y, Z reference vA = velocity of the origin A of the x, y, z reference, measured from the X, Y, Z reference (vB>A)xyz = velocity of “B with respect to A,” as measured by an observer attached to the rotating x, y, z reference  = angular velocity of the x, y, z reference, measured from the X, Y, Z reference rB>A = position of B with respect to A

16

Comparing Eq. 16–26 with Eq. 16–16 (vB = vA +  * rB>A), which is valid for a translating frame of reference, it can be seen that the only difference between these two equations is represented by the term (vB>A)xyz . When applying Eq. 16–26 it is often useful to understand what each of the terms represents. In order of appearance, they are as follows:

vB

e

absolute velocity of B

f

motion of B observed from the X, Y, Z frame

(equals)

vA

e

absolute velocity of the origin of x, y, z frame (plus)

motion of x, y, z frame

y observed from the  * rB>A e

(vB>A)xyz

e

X, Y, Z frame

angular velocity effect caused by rotation of x, y, z frame (plus) velocity of B with respect to A

f

motion of B observed from the x, y, z frame

392

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

Acceleration.

The acceleration of B, observed from the X, Y, Z coordinate system, may be expressed in terms of its motion measured with respect to the rotating system of coordinates by taking the time derivative of Eq. 16–26. drB>A d(vB>A)xyz dvB dvA d = + * rB>A +  * + dt dt dt dt dt dr d(v ) # B>A B>A xyz aB = aA +  * rB>A +  * + dt dt

(16–27)

# Here  = d >dt is the angular acceleration of the x, y, z coordinate # system. Since  is always perpendicular to the plane of motion, then  measures only the change in magnitude of . The derivative drB>A >dt is defined by Eq. 16–25, so that  * 16

drB>A dt

=  * (vB>A)xyz +  * ( * rB>A)

(16–28)

Finding the time derivative of (vB>A)xyz = (vB>A)xi + (vB>A)y j, d(vB>A)xyz dt

= c

d(vB>A)x dt

i +

d(vB>A)y dt

j d + c (vB>A)x

dj di + (vB>A)y d dt dt

The two terms in the first set of brackets represent the components of acceleration of point B as measured by an observer attached to the rotating coordinate system. These terms will be denoted by (aB>A)xyz . The terms in the second set of brackets can be simplified using Eqs. 16–24. d(vB>A)xyz dt

= (aB>A)xyz +  * (vB>A)xyz

Substituting this and Eq. 16–28 into Eq. 16–27 and rearranging terms, # aB = aA +  * rB>A +  * ( * rB>A) + 2 * (vB>A)xyz + (aB>A)xyz (16–29) where aB = acceleration of B, measured from the X, Y, Z reference aA = acceleration of the origin A of the x, y, z reference, measured from the X, Y, Z reference (aB>A)xyz , (vB>A)xyz = acceleration and velocity of B with respect to A, as measured by an observer attached to the rotating x, y, z reference #  ,  = angular acceleration and angular velocity of the x, y, z reference, measured from the X, Y, Z reference rB>A = position of B with respect to A

16.8

relatIVe-motIon analysIs usIng rotatIng axes

393

If Eq. 16–29 is compared with Eq. 16–18, written in the form # aB = aA +  * rB>A +  * ( * rB>A), which is valid for a translating frame of reference, it can be seen that the difference between these two equations is represented by the terms 2 * (vB>A)xyz and (aB>A)xyz . In particular, 2 * (vB>A)xyz is called the Coriolis acceleration, named after the French engineer G. C. Coriolis, who was the first to determine it. This term represents the difference in the acceleration of B as measured from nonrotating and rotating x, y, z axes. As indicated by the vector cross product, the Coriolis acceleration will always be perpendicular to both  and (vB>A)xyz . It is an important component of the acceleration which must be considered whenever rotating reference frames are used. This often occurs, for example, when studying the accelerations and forces which act on rockets, long-range projectiles, or other bodies having motions whose measurements are significantly affected by the rotation of the earth. The following interpretation of the terms in Eq. 16–29 may be useful when applying this equation to the solution of problems.

aB

e absolute acceleration of B (equals)

aA

#  * rB>A

e

2 * (vB>A)xyz

absolute acceleration of the origin of x, y, z frame (plus)

angular acceleration effect c caused by rotation of x, y, z frame (plus)

 * ( * rB>A) e

motion of B observed f from the X, Y, Z frame

motion of x, y, z frame y observed from the X, Y, Z frame

angular velocity effect caused by rotation of x, y, z frame (plus)

combined effect of B moving c relative to x, y, z coordinates sinteracting motion and rotation of x, y, z frame (plus)

(aB>A)xyz

e

acceleration of B with motion of B observed f respect to A from the x, y, z frame

16

394

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

Procedure for Analysis Equations 16–26 and 16–29 can be applied to the solution of problems involving the planar motion of particles or rigid bodies using the following procedure. coordinate Axes. • Choose an appropriate location for the origin and proper orientation of the axes for both fixed X, Y, Z and moving x, y, z reference frames.

• Most often solutions are easily obtained if at the instant considered: 1. the origins are coincident 2. the corresponding axes are collinear 3. the corresponding axes are parallel

• The moving frame should be selected fixed to the body or device

16

along which the relative motion occurs.

Kinematic Equations. • After defining the origin A of the moving reference and specifying the moving point B, Eqs. 16–26 and 16–29 should be written in symbolic form vB = vA +  * rB>A + (vB>A)xyz # aB = aA +  * rB>A +  * ( * rB>A) + 2 * (vB>A)xyz + (aB>A)xyz

• The Cartesian components of all these vectors may be expressed

along either the X, Y, Z axes or the x, y, z axes. The choice is arbitrary provided a consistent set of unit vectors is used.

of the moving reference is expressed by vA , aA , , and • Motion #

 ; and motion of B with respect to the moving reference is expressed by rB>A , (vB>A)xyz , and (aB>A)xyz .

y B x

C A

The rotation of the dumping bin of the truck about point C is operated by the extension of the hydraulic cylinder AB. To determine the rotation of the bin due to this extension, we can use the equations of relative motion and fix the x, y axes to the cylinder so that the relative motion of the cylinder’s extension occurs along the y axis.

16.8

EXAMPLE

relatIVe-motIon analysIs usIng rotatIng axes

395

16.18

At the instant u = 60, the rod in Fig. 16–33 has an angular velocity of 3 rad>s and an angular acceleration of 2 rad>s2. At this same instant, collar C travels outward along the rod such that when x = 0.2 m the velocity is 2 m>s and the acceleration is 3 m>s2, both measured relative to the rod. Determine the Coriolis acceleration and the velocity and acceleration of the collar at this instant. SoLution Coordinate Axes. The origin of both coordinate systems is located at point O, Fig. 16–33. Since motion of the collar is reported relative to the rod, the moving x, y, z frame of reference is attached to the rod. Kinematic equations. vC = vO +  * rC>O + (vC>O)xyz (1) # aC = aO +  * rC>O +  * ( * rC>O) + 2 * (vC>O)xyz + (aC>O)xyz (2) It will be simpler to express the data in terms of i, j, k component vectors rather than I, J, K components. Hence, Motion of moving reference vO = 0 aO = 0  = 5 -3k 6 rad>s #  = 5 -2k 6 rad>s2

Y y x  0.2 m

O 2

Motion of C with respect to moving reference rC>O = 5 0.2i 6 m

rad/s2 C 30

x

Fig. 16–33

(aC>O)xyz = 5 3i 6 m>s2 Ans.

This vector is shown dashed in Fig. 16–33. If desired, it may be resolved into I, J components acting along the X and Y axes, respectively. The velocity and acceleration of the collar are determined by substituting the data into Eqs. 1 and 2 and evaluating the cross products, which yields vC = vO +  * rC>O + (vC>O)xyz = 0 + (-3k) * (0.2i) + 2i = 5 2i - 0.6j 6 m>s # aC = aO +  * rC>O +  * ( * rC>O) + 2 * (vC>O)xyz + (aC>O)xyz

Ans.

= 0 + (-2k) * (0.2i) + (-3k) * [(-3k) * (0.2i)] + 2(-3k) * (2i) + 3i = 0 - 0.4j - 1.80i - 12j + 3i = 5 1.20i - 12.4j 6 m>s2

3 m/s2 2 m/s

aCor

(vC>O)xyz = 5 2i 6 m>s

The Coriolis acceleration is defined as aCor = 2 * (vC>O)xyz = 2(-3k) * (2i) = 5 -12j 6 m>s2

16

u  60

3 rad/s

Ans.

X

396

Chapter 16

EXAMPLE

p l a n a r K I n e m at I C s

rIgId Body

16.19 Rod AB, shown in Fig. 16–34, rotates clockwise such that it has an angular velocity vAB = 3 rad>s and angular acceleration aAB = 4 rad>s2 when u = 45. Determine the angular motion of rod DE at this instant. The collar at C is pin connected to AB and slides over rod DE.

Y, y B 0.4 m D

VDE, ADE 0.4 m

of a

C

vAB  3 rad/s aAB  4 rad/s2 u  45 A

E

X, x

SoLution Coordinate Axes. The origin of both the fixed and moving frames of reference is located at D, Fig. 16–34. Furthermore, the x, y, z reference is attached to and rotates with rod DE so that the relative motion of the collar is easy to follow. Kinematic equations.

Fig. 16–34

vC = vD +  * rC>D + (vC>D)xyz (1) # aC = aD +  * rC>D +  * ( * rC>D) + 2 * (vC>D)xyz + (aC>D)xyz (2)

16

All vectors will be expressed in terms of i, j, k components. Motion of moving reference vD aD  # 

= 0 = 0 = -vDEk

Motion of C with respect to moving reference rC>D = 5 0.4i 6 m (vC>D)xyz = (vC>D)xyzi (aC>D)xyz = (aC>D)xyzi

= -aDEk

motion of C: Since the collar moves along a circular path of radius AC, its velocity and acceleration can be determined using Eqs. 16–9 and 16–14. vC = VAB * rC>A = ( -3k) * (0.4i + 0.4j) = 5 1.2i - 1.2j 6 m>s aC = AAB * rC>A - v2ABrC>A = ( -4k) * (0.4i + 0.4j) - (3)2(0.4i + 0.4j) = 5 -2i - 5.2j 6 m>s2

Substituting the data into Eqs. 1 and 2, we have vC = vD +  * rC>D + (vC>D)xyz 1.2i - 1.2j = 0 + ( -vDEk) * (0.4i) + (vC>D)xyzi 1.2i - 1.2j = 0 - 0.4vDE j + (vC>D)xyzi (vC>D)xyz = 1.2 m>s vDE = 3 rad>s b Ans. # aC = aD +  * rC>D +  * ( * rC>D) + 2 * (vC>D)xyz + (aC>D)xyz -2i - 5.2j = 0 + (-aDEk) * (0.4i) + (-3k) * [( -3k) * (0.4i)] + 2(-3k) * (1.2i) + (aC>D)xyzi -2i - 5.2j = -0.4aDE j - 3.6i - 7.2j + (aC>D)xyzi (aC>D)xyz = 1.6 m>s2 aDE = -5 rad>s2 = 5 rad>s2d Ans.

16.8

EXAMPLE

397

relatIVe-motIon analysIs usIng rotatIng axes

16.20

Planes A and B fly at the same elevation and have the motions shown in Fig. 16–35. Determine the velocity and acceleration of A as measured by the pilot of B. SoLution Coordinate Axes. Since the relative motion of A with respect to the pilot in B is being sought, the x, y, z axes are attached to plane B, Fig. 16–35. At the instant considered, the origin B coincides with the origin of the fixed X, Y, Z frame. Kinematic equations. vA = vB +  * rA>B + (vA>B)xyz (1) # aA = aB +  * rA>B +  * ( * rA>B) + 2 * (vA>B)xyz + (aA>B)xyz (2) Motion of Moving reference: vB = 5 600j 6 km>h

700 km/h

v2B (600)2 = = 900 km>h2 r 400 = (aB)n + (aB)t = 5 900i - 100j 6 km>h2 50 km/h2 600 km>h vB =  = 5 -1.5k 6 rad>h = = 1.5 rad>h b r 400 km # 100 km>h2 (aB)t = = = 0.25 rad>h2d  = 5 0.25k 6 rad>h2 r 400 km

(aB)n = aB  # 

y, Y

Motion of A with respect to Moving reference:

rA>B = 5 -4i 6 km (vA>B)xyz = ? (aA>B)xyz = ?

Substituting the data into Eqs. 1 and 2, realizing that vA = 5 700j 6 km>h and aA = 5 50j 6 km>h2, we have vA = vB +  * rA>B + (vA>B)xyz 700j = 600j + (-1.5k) * (-4i) + (vA>B)xyz Ans. (vA>B)xyz = 5 94j 6 km>h # aA = aB +  * rA>B +  * ( * rA>B) + 2 * (vA>B)xyz + (aA>B)xyz 50j = (900i - 100j) + (0.25k) * (-4i) + (-1.5k) * [(-1.5k) * (-4i)] + 2( -1.5k) * (94j) + (aA>B)xyz (aA>B)xyz = 5 -1191i + 151j 6 km>h2 Ans.

Note: The solution of this problem should be compared with that of

Example 12.26, where it is seen that (vB>A)xyz  (vA>B)xyz and (aB>A)xyz  (aA>B)xyz.

600 km/h A

rA/B

100 km/h2 x, X B

400 km 4 km

Fig. 16–35

16

398

Chapter 16

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of a

rIgId Body

Problems 16–129. At the instant shown, ball B is rolling along the slot in the disk with a velocity of 600 mm>s and an acceleration of 150 mm>s2, both measured relative to the disk and directed away from O. If at the same instant the disk has the angular velocity and angular acceleration shown, determine the velocity and acceleration of the ball at this instant.

16–131. Ball C moves with a speed of 3 m>s, which is increasing at a constant rate of 1.5 m>s2, both measured relative to the circular plate and directed as shown. At the same instant the plate rotates with the angular velocity and angular acceleration shown. Determine the velocity and acceleration of the ball at this instant. z v a

8 rad/s 5 rad/s2

z v  6 rad/s a  3 rad/s2 O

16 0.8 m

B

O

300 mm

0.4 m

C

x

x

y

y

Prob. 16–129

Prob. 16–131

16–130. Block A, which is attached to a cord, moves along the slot of a horizontal forked rod. At the instant shown, the cord is pulled down through the hole at O with an acceleration of 4 m>s2 and its velocity is 2 m>s. Determine the acceleration of the block at this instant. The rod rotates about O with a constant angular velocity v = 4 rad>s.

*16–132. Particles B and A move along the parabolic and circular paths, respectively. If B has a velocity of 7 m>s in the direction shown and its speed is increasing at 4 m>s2, while A has a velocity of 8 m>s in the direction shown and its speed is decreasing at 6 m>s2, determine the relative velocity and relative acceleration of B with respect to A. y y  x2

B vB  7 m/ s y

x

2m vA  8 m/ s

A A v

1m

O 100 mm

Prob. 16–130

Prob. 16–132

x

16.8 16–133. Rod AB rotates counterclockwise with a constant angular velocity v = 3 rad>s. Determine the velocity of point C located on the double collar when u = 30°. The collar consists of two pin-connected slider blocks which are constrained to move along the circular path and the rod AB. 16–134. Rod AB rotates counterclockwise with a constant angular velocity v = 3 rad>s. Determine the velocity and acceleration of point C located on the double collar when u = 45°. The collar consists of two pin-connected slider blocks which are constrained to move along the circular path and the rod AB.

*16–136. If the piston is moving with a velocity yA = 3 m>s and acceleration of aA = 1.5 m>s2, determine the angular velocity and angular acceleration of the slotted link at the instant shown. Link AB slides freely along its slot on the fixed peg C.

3 m/s 1.5 m/s2

vA

aA

B

C

399

relatIVe-motIon analysIs usIng rotatIng axes

A

30 C 0.5 m B

v = 3 rad/s u

Prob. 16–136

A

16

0.4 m

Probs. 16–133/134 16–135. A girl stands at A on a platform which is rotating with an angular acceleration a = 0.2 rad>s2 and at the instant shown has an angular velocity v = 0.5 rad>s. If she walks at a constant speed y = 0.75 m>s measured relative to the platform, determine her acceleration (a) when she reaches point D in going along the path ADC, d = 1 m; and (b) when she reaches point B if she follows the path ABC, r = 3 m.

16–137. Water leaves the impeller of the centrifugal pump with a velocity of 25 m>s and acceleration of 30 m>s2, both measured relative to the impeller along the blade line AB. Determine the velocity and acceleration of a water particle at A as it leaves the impeller at the instant shown. The impeller rotates with a constant angular velocity of v = 15 rad>s.

y B 30

O A

d r

A

y x

C

D

v  15 rad/s 0.3 m

B x

Prob. 16–135

Prob. 16–137

400

Chapter 16

p l a n a r K I n e m at I C s

of a

16–138. Peg B on the gear slides freely along the slot in link AB. If the gear’s center O moves with the velocity and acceleration shown, determine the angular velocity and angular acceleration of the link at this instant.

rIgId Body 16–141. At the instant shown rod AB has an angular velocity vAB = 4 rad>s and an angular acceleration aAB = 2 rad>s2. Determine the angular velocity and angular acceleration of rod CD at this instant. The collar at C is pin connected to CD and slides freely along AB.

B

150 mm vO  3 m/s aO  1.5 m/s2

vAB  4 rad/s aAB  2 rad/s2

A

600 mm O

60

150 mm

A

0.5 m

0.75 m

C

D

Prob. 16–138

16

B

16–139. The collar C is pinned to rod CD while it slides on rod AB. If rod AB has an angular velocity of 2 rad>s and an angular acceleration of 8 rad>s2, both acting counterclockwise, determine the angular velocity and the angular acceleration of rod CD at the instant shown. vAB  2 rad/s aAB  8 rad/s2

A

Prob. 16–141 16–142. Collar B moves to the left with a speed of 5 m>s, which is increasing at a constant rate of 1.5 m>s2, relative to the hoop, while the hoop rotates with the angular velocity and angular acceleration shown. Determine the magnitudes of the velocity and acceleration of the collar at this instant. v  6 rad/s a  3 rad/s2

A

60 C 1.5 m

B 1m

D

450 mm 200 mm

Prob. 16–139

B

*16–140. At the instant shown, the robotic arm AB is rotating counterclockwise at v = 5 rad>s and has an angular acceleration a = 2 rad>s2. Simultaneously, the grip BC is rotating counterclockwise at v = 6 rad>s and a = 2 rad>s2, both measured relative to a fixed reference. Determine the velocity and acceleration of the object held at the grip C.

Prob. 16–142 16–143. Block D of the mechanism is confined to move within the slot of member CB. If link AD is rotating at a constant rate of vAD = 4 rad>s, determine the angular velocity and angular acceleration of member CB at the instant shown.

125 mm

y

B

C

B

15

300 mm

D

300 mm

v¿, a¿

30

A

v, a

Prob. 16–140

200 mm vAD  4 rad/s

30

x C

A

Prob. 16–143

16.8 *16–144. A ride in an amusement park consists of a rotating arm AB having a constant angular velocity vAB = 2 rad>s about point A and a car mounted at the end of the arm which  has a constant angular velocity V = {−0.5k} rad>s, measured relative to the arm. At the instant shown, determine the velocity and acceleration of the passenger at C. 16–145. A ride in an amusement park consists of a rotating arm AB that has an angular acceleration of aAB = 1 rad>s2 when vAB = 2 rad>s at the instant shown. Also at this instant the car mounted at the end of the arm has a relative angular acceleration of A = {−0.6k} rad>s2 when V = {−0.5k} rad>s. Determine the velocity and acceleration of the passenger C at this instant.

relatIVe-motIon analysIs usIng rotatIng axes

16–147. At the instant shown, car A travels with a speed of 25 m>s, which is decreasing at a constant rate of 2 m>s2, while car C travels with a speed of 15 m>s, which is increasing at a constant rate of 3 m>s2. Determine the velocity and acceleration of car A with respect to car C.

45

250 m

15 m/s 2 m/s2

C

v¿  0.5 rad/s B B 5m

y

30

16

2

2 m/s

C

Prob. 16–147

x

A

25 m/s

A

vAB  2 rad/s

15 m/s 3 m/s2

200 m 1m

60

401

Probs. 16–144/145 16–146. If the slider block C is fixed to the disk that has a constant counterclockwise angular velocity of 4 rad>s, determine the angular velocity and angular acceleration of the slotted arm AB at the instant shown.

40 mm C

*16–148. At the instant shown, car B travels with a speed of 15 m>s, which is increasing at a constant rate of 2 m>s2, while car C travels with a speed of 15 m>s, which is increasing at a constant rate of 3 m>s2. Determine the velocity and acceleration of car B with respect to car C.

B

60 mm 30 v  4 rad/s

45

250 m

15 m/s 2 m/s2

180 mm

C

B 200 m A

60

25 m/s 2 m/s2

A

Prob. 16–146

Prob. 16–148

15 m/s 3 m/s2

402

Chapter 16

p l a n a r K I n e m at I C s

of a

16–149. A ride in an amusement park consists of a rotating platform P, having constant angular velocity vP = 1.5 rad>s, and four cars, C, mounted on the platform, which have constant angular velocities vC>P = 2 rad>s measured relative to the platform. Determine the velocity and acceleration of the passenger at B at the instant shown.

rIgId Body 16–151. The disk rotates with the angular motion shown. Determine the angular velocity and angular acceleration of the slotted link AC at this instant. The peg at B is fixed to the disk.

A

Y

0.75 m

30

A 0.75 m vC /P

vP C

2 rad/s

3m

0.3 m

C 1.5 rad/s P C

B

30

X

C

B

v  6 rad/s a  10 rad/s2

16

C

Prob. 16–151

Prob. 16–149

16–150. The two-link mechanism serves to amplify angular motion. Link AB has a pin at B which is confined to move within the slot of link CD. If at the instant shown, AB (input) has an angular velocity of vAB = 2.5 rad>s, determine the angular velocity of CD (output) at this instant.

*16–152. The “quick-return” mechanism consists of a crank AB, slider block B, and slotted link CD. If the crank has the angular motion shown, determine the angular motion of the slotted link at this instant.

D

B 100 mm

D

150 mm

vAB aAB

3 rad/s 9 rad/s2

C

30 A

30 45

A 30

vAB  2.5 rad/s

vCD, aCD

Prob. 16–150

300 mm

C

Prob. 16–152

B

16.8

relatIVe-motIon analysIs usIng rotatIng axes

403

ConCePtuAl Problems C16–1. An electric motor turns the tire at A at a constant angular velocity, and friction then causes the tire to roll without slipping on the inside rim of the Ferris wheel. Using appropriate numerical values, determine the magnitude of the velocity and acceleration of passengers in one of the baskets. Do passengers in the other baskets experience this same motion? Explain.

C16–3. The bi-fold hangar door is opened by cables that move upward at a constant speed of 0.5 m>s. Determine the angular velocity of BC and the angular velocity of AB when u = 45. Panel BC is pinned at C and has a height which is the same as the height of BA. Use appropriate numerical values to explain your result.

C

A

16

u B

A

Prob. C16–3 Prob. C16–1 C16–2. The crank AB turns counterclockwise at a constant rate V causing the connecting arm CD and rocking beam DE to move. Draw a sketch showing the location of the IC for the connecting arm when u = 0, 90, 180, and 270. Also, how was the curvature of the head at E determined, and why is it curved in this way?

E

C16–4. If the tires do not slip on the pavement, determine the points on the tire that have a maximum and minimum speed and the points that have a maximum and minimum acceleration. Use appropriate numerical values for the car’s speed and tire size to explain your result.

D

C A

Prob. C16–2

B

u

Prob. C16–4

404

Chapter 16

p l a n a r K I n e m at I C s

of a

rIgId Body

CHAPter reVieW Rigid-Body Planar Motion A rigid body undergoes three types of planar motion: translation, rotation about a fixed axis, and general plane motion. Path of rectilinear translation

Translation When a body has rectilinear translation, all the particles of the body travel along parallel straight-line paths. If the paths have the same radius of curvature, then curvilinear translation occurs. Provided we know the motion of one of the particles, then the motion of all of the others is also known. 16

Path of curvilinear translation

Rotation about a Fixed Axis For this type of motion, all of the particles move along circular paths. Here, all line segments in the body undergo the same angular displacement, angular velocity, and angular acceleration. Once the angular motion of the body is known, then the velocity of any particle a distance r from the axis can be obtained. The acceleration of any particle has two components. The tangential component accounts for the change in the magnitude of the velocity, and the normal component accounts for the change in the velocity’s direction.

Rotation about a fixed axis

v = v 0 + act

v = du>dt a = dv>dt a du = v dv

or

u = u0 + v0t + 12 act2 v2 = v20 + 2ac(u - u0) Constant ac

v = vr

at = ar, an = v2r

General Plane Motion When a body undergoes general plane motion, it simultaneously translates and rotates. There are several methods for analyzing this motion. Absolute Motion Analysis If the motion of a point on a body or the angular motion of a line is known, then it may be possible to relate this motion to that of another point or line using an absolute motion analysis. To do so, linear position coordinates s or angular position coordinates u are established (measured from a fixed point or line). These position coordinates are then related using the geometry of the body. The time derivative of this equation gives the relationship between the velocities and/or the angular velocities. A second time derivative relates the accelerations and/or the angular accelerations.

General plane motion

Chapter reVIew

Relative-Motion using Translating Axes General plane motion can also be analyzed using a relative-motion analysis between two points A and B located on the body. This method considers the motion in parts: first a translation of the selected base point A, then a relative “rotation” of the body about point A, which is measured from a translating axis. Since the relative motion is viewed as circular motion about the base point, point B will have a velocity vB>A that is tangent to the circle. It also has two components of acceleration, (aB>A)t and (aB>A)n . It is also important to realize that aA and aB will have tangential and normal components if these points move along curved paths.

vB = vA + V * rB>A aB = aA + A * rB>A - v2 rB>A

Instantaneous Center of Zero Velocity If the base point A is selected as having zero velocity, then the relative velocity equation becomes vB = V * rB>A . In this case, motion appears as if the body rotates about an instantaneous axis passing through A. The instantaneous center of rotation (IC) can be established provided the directions of the velocities of any two points on the body are known, or the velocity of a point and the angular velocity are known. Since a radial line r will always be perpendicular to each velocity, then the IC is at the point of intersection of these two radial lines. Its measured location is determined from the geometry of the body. Once it is established, then the velocity of any point P on the body can be determined from v = vr, where r extends from the IC to point P.

Relative Motion using Rotating Axes Problems that involve connected members that slide relative to one another or points not located on the same body can be analyzed using a relative-motion analysis referenced from a rotating frame. This gives rise to the term 2 * (vB>A)xyz that is called the Coriolis acceleration.

405

16

A

rA/IC P

V

rP/IC

vA vP B

vIC  0 IC

rB/IC

vB

vB = vA +  * rB>A + (vB>A)xyz # aB = aA +  * rB>A +  * ( * rB>A) + 2 * (vB>A)xyz + (aB>A)xyz

406

Chapter 16

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of a

rIgId Body

reVieW Problems R16–1. The hoisting gear A has an initial angular velocity of 60 rad>s and a constant deceleration of 1 rad>s2. Determine the velocity and deceleration of the block which is being hoisted by the hub on gear B when t = 3 s. v, a 0.15 m

0.6 m

0.3 m A

R16–3. The board rests on the surface of two drums. At the instant shown, it has an acceleration of 0.5 m>s2 to the right, while at the same instant, points on the outer rim of each drum have an acceleration with a magnitude of 3 m>s2. If the board does not slip on the drums, determine its speed due to the motion.

B

a  0.5 m/s2 250 mm

250 mm

16

Prob. R16–3

Prob. R16–1 R16–2. Starting at (vA )0 = 3 rad>s, when u = 0, s = 0, pulley A is given an angular acceleration a = (0.6u) rad>s2, where u is in radians. Determine the speed of block B when it has risen s = 0.5 m. The pulley has an inner hub D which is fixed to C and turns with it. 150 mm A 50 mm

C

D

R16–4. If bar AB has an angular velocity vAB = 6 rad>s, determine the velocity of the slider block C at the instant shown.

75 mm

vAB  6 rad/s B s

Prob. R16–2

B

200 mm A

u  45

500 mm 30

Prob. R16–4

C

407

reVIew proBlems R16–5. The center of the pulley is being lifted vertically with an acceleration of 4 m>s2 at the instant it has a velocity of 2 m>s. If the cable does not slip on the pulley’s surface, determine the accelerations of the cylinder B and point C on the pulley.

R16–7. The disk is moving to the left such that it has an angular acceleration a = 8 rad>s2 and angular velocity v = 3 rad>s at the instant shown. If it does not slip at A, determine the acceleration of point B.

aA = 4 m/s2 vA = 2 m/s v  3 rad/s a  8 rad/s2 C

D

A

30

80 mm

B

0.5 m

C

A

Prob. R16–7 B

16

Prob. R16–5 R16–6. At the instant shown, link AB has an angular velocity vAB = 2 rad>s and an angular acceleration aAB = 6 rad>s2. Determine the acceleration of the pin at C and the angular acceleration of link CB at this instant, when u = 60°.

R16–8. At the given instant member AB has the angular motions shown. Determine the velocity and acceleration of the slider block C at this instant.

B

300 mm

A

0.7 m

B vAB  2 rad/s aAB  6 rad/s2

A

3 rad/s 2 rad/s2

500 mm 0.5 m

u C

C

D 5

175 mm

Prob. R16–6

0.5 m 3

4

Prob. R16–8

Chapter 17

(© Surasaki/Fotolia) Tractors and other heavy equipment can be subjected to severe loadings due to dynamic loadings as they accelerate. In this chapter we will show how to determine these loadings for planar motion.

Planar Kinetics of a Rigid Body: Force and Acceleration Chapter ObjeCtives n

To introduce the methods used to determine the mass moment of inertia of a body.

n

To develop the planar kinetic equations of motion for a symmetric rigid body.

n

To discuss applications of these equations to bodies undergoing translation, rotation about a fixed axis, and general plane motion.

17.1

Mass Moment of Inertia

Since a body has a definite size and shape, an applied nonconcurrent force system can cause the body to both translate and rotate. The translational aspects of the motion were studied in Chapter 13 and are governed by the equation F = ma. It will be shown in the next section that the rotational aspects, caused by a moment M, are governed by an equation of the form M = IA. The symbol I in this equation is termed the mass moment of inertia. By comparison, the moment of inertia is a measure of the resistance of a body to angular acceleration (M = IA) in the same way that mass is a measure of the body’s resistance to acceleration (F = ma).

Video Solutions are available for selected questions in this chapter.

410

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

The flywheel on the engine of this tractor has a large moment of inertia about its axis of rotation. Once it is set into motion, it will be difficult to stop, and this in turn will prevent the engine from stalling and instead will allow it to maintain a constant power.

We define the moment of inertia as the integral of the “second moment” about an axis of all the elements of mass dm which compose the body.* For example, the body’s moment of inertia about the z axis in Fig. 17–1 is

z

17

I =

r

dm

Fig. 17–1

Lm

r 2 dm

(17–1)

Here the “moment arm” r is the perpendicular distance from the z axis to the arbitrary element dm. Since the formulation involves r, the value of I is different for each axis about which it is computed. In the study of planar kinetics, the axis chosen for analysis generally passes through the body’s mass center G and is always perpendicular to the plane of motion. The moment of inertia about this axis will be denoted as IG . Since r is squared in Eq. 17–1, the mass moment of inertia is always a positive quantity. Common units used for its measurement are kg # m2. If the body consists of material having a variable density, r = r (x,y,z), the elemental mass dm of the body can be expressed in terms of its density and volume as dm = r dV. Substituting dm into Eq. 17–1, the body’s moment of inertia is then computed using volume elements for integration; i.e., I =

LV

r2r dV

(17–2)

*Another property of the body, which measures the symmetry of the body’s mass with respect to a coordinate system, is the product of inertia. This property applies to the threedimensional motion of a body and will be discussed in Chapter 21.

17.1

In the special case of r being a constant, this term may be factored out of the integral, and the integration is then purely a function of geometry, I = r

LV

r2 dV

411

Mass MoMent of inertia z

dm  r dV

(17–3) z

When the volume element chosen for integration has infinitesimal dimensions in all three directions, Fig. 17–2a, the moment of inertia of the body must be determined using “triple integration.” The integration process can, however, be simplified to a single integration provided the chosen volume element has a differential size or thickness in only one direction. Shell or disk elements are often used for this purpose.

y x

y x (a)

z

procedure for analysis z

To obtain the moment of inertia by integration, we will consider only symmetric bodies having volumes which are generated by revolving a curve about an axis. An example of such a body is shown in Fig. 17–2a. Two types of differential elements can be chosen.

y y

Shell Element. • If a shell element having a height z, radius r = y, and thickness dy is chosen for integration, Fig. 17–2b, then the volume is dV = (2py)(z)dy.

(b)

z

• This element may be used in Eq. 17–2 or 17–3 for determining the

y

moment of inertia Iz of the body about the z axis, since the entire element, due to its “thinness,” lies at the same perpendicular distance r = y from the z axis (see Example 17.1).

dz

Disk Element. • If a disk element having a radius y and a thickness dz is chosen for integration, Fig. 17–2c, then the volume is dV = (py2)dz.

z y

• This element is finite in the radial direction, and consequently its

parts do not all lie at the same radial distance r from the z axis. As a result, Eq. 17–2 or 17–3 cannot be used to determine Iz directly. Instead, to perform the integration it is first necessary to determine the moment of inertia of the element about the z axis and then integrate this result (see Example 17.2).

dy

x

x (c)

Fig. 17–2

17

412

Chapter 17

ExamplE

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

17.1 Determine the moment of inertia of the cylinder shown in Fig. 17–3a about the z axis. The density of the material, r, is constant. z

z

r

dr

R h 2

h 2

O

y

O

h 2

x

y h 2

x

(a)

(b)

Fig. 17–3

17

Solution shell element. This problem can be solved using the shell element in Fig. 17–3b and a single integration. The volume of the element is dV = (2pr)(h) dr, so that its mass is dm = rdV = r(2phr dr). Since the entire element lies at the same distance r from the z axis, the moment of inertia of the element is dIz = r2dm = r2phr3 dr Integrating over the entire region of the cylinder yields Iz =

Lm

2

r dm = r2ph

L0

R

r3 dr =

rp 4 Rh 2

The mass of the cylinder is m =

Lm

dm = r2ph

L0

R

r dr = rphR2

so that Iz =

1 mR2 2

Ans.

17.1

Mass MoMent of inertia

413

17.2

ExamplE

If the density of the material is 3000 kg>m3, determine the moment of inertia of the solid in Fig. 17–4a about the y axis.

y y x

1m

1m

dy 1m

1m

y2  x

(x,y)

x (a)

y x

(b)

Fig. 17–4

Solution Disk element. The moment of inertia will be found using a disk element, as shown in Fig. 17–4b. Here the element intersects the curve at the arbitrary point (x,y) and has a mass dm = r dV = r(px2) dy Although all portions of the element are not located at the same distance from the y axis, it is still possible to determine the moment of inertia dIy of the element about the y axis. In the preceding example it was shown that the moment of inertia of a cylinder about its longitudinal axis is I = 12 mR2, where m and R are the mass and radius of the cylinder. Since the height is not involved in this formula, the disk itself can be thought of as a cylinder. Thus, for the disk element in Fig. 17–4b, we have dIy = 12(dm)x2 = 12[r(px2) dy]x2 Substituting x = y2, r = 3000 kg>m3, and integrating with respect to y, from y = 0 to y = 1 m, yields the moment of inertia for the entire solid. 3

Iy =

p(3000 kg>m ) 2

L0

1m

x4 dy =

p(3000) 1 m 8 y dy = 524 kg # m2 Ans. 2 L0

17

414

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n y¿

dm r

r¿ G

A

x¿ x¿

d

z

y¿

z¿

Fig. 17–5

17

Parallel-Axis Theorem. If the moment of inertia of the body about an axis passing through the body’s mass center is known, then the moment of inertia about any other parallel axis can be determined by using the parallel-axis theorem. This theorem can be derived by considering the body shown in Fig. 17–5. Here the z axis passes through the mass center G, whereas the corresponding parallel z axis lies at a constant distance d away. Selecting the differential element of mass dm, which is located at point (x, y), and using the Pythagorean theorem, r2 = (d + x)2 + y2, we can express the moment of inertia of the body about the z axis as I =

Lm

r2 dm = =

[(d + x)2 + y2] dm Lm (x2 + y2) dm + 2d x dm + d2 dm Lm Lm Lm

Since r2 = x2 + y2, the first integral represents IG . The second integral equals zero, since the z axis passes through the body’s mass center, i.e., 1 xdm = xm = 0 since x = 0. Finally, the third integral

17.1

Mass MoMent of inertia

415

represents the total mass m of the body. Hence, the moment of inertia about the z axis can be written as I = IG + md 2

(17–4)

where IG = moment of inertia about the z axis passing through the mass center G m = mass of the body d = perpendicular distance between the parallel z and z axes

Radius of Gyration.

Occasionally, the moment of inertia of a body about a specified axis is reported in handbooks using the radius of gyration, k. This is a geometrical property which has units of length. When it and the body’s mass m are known, the body’s moment of inertia is determined from the equation

I = mk2 or k =

I Am

(17–5)

Note the similarity between the definition of k in this formula and r in the equation dI = r2 dm, which defines the moment of inertia of an elemental mass dm of the body about an axis.

Composite Bodies. If a body consists of a number of simple

shapes such as disks, spheres, and rods, the moment of inertia of the body about any axis can be determined by adding algebraically the moments of inertia of all the composite shapes computed about the axis. Algebraic addition is necessary since a composite part must be considered as a negative quantity if it has already been counted as a piece of another part—for example, a “hole” subtracted from a solid plate. The parallelaxis theorem is needed for the calculations if the center of mass of each composite part does not lie on the axis. For the calculation, then, I = (IG + md2). Here IG for each of the composite parts is determined by integration, or for simple shapes, such as rods and disks, it can be found from a table, such as the one given on the inside back cover of this book.

17

416

Chapter 17

ExamplE

planar KinetiCs

of a

rigid Body: forCe

a C C e l e r at i o n

and

17.3 If the plate shown in Fig. 17–6a has a density of 8000 kg>m3 and a thickness of 10 mm, determine its moment of inertia about an axis directed perpendicular to the page and passing through point O.

250 mm

125 mm

G

G

250 mm O



G

125 mm

Thickness 10 mm (b)

(a)

Fig. 17–6

17

Solution The plate consists of two composite parts, the 250-mm-radius disk minus a 125-mm-radius disk, Fig. 17–6b. The moment of inertia about O can be determined by computing the moment of inertia of each of these parts about O and then adding the results algebraically. The calculations are performed by using the parallel-axis theorem in conjunction with the data listed in the table on the inside back cover. Disk. The moment of inertia of a disk about the centroidal axis perpendicular to the plane of the disk is IG = 12 mr2. The mass center of the disk is located at a distance of 0.25 m from point O. Thus, md = rdVd = 8000 kg>m3 [p(0.25 m)2(0.01 m)] = 15.71 kg (Id)O = 12 mdr2d + mdd2 1 = (15.71 kg)(0.25 m)2 + (15.71 kg)(0.25 m)2 2 = 1.473 kg # m2 hole. For the 125-mm-radius disk (hole), we have mh = rhVh = 8000 kg>m3 [p(0.125 m)2(0.01 m)] = 3.927 kg (Ih)O = 12 mhr2h + mhd2 1 = (3.927 kg)(0.125 m)2 + (3.927 kg)(0.25 m)2 2 = 0.276 kg # m2 The moment of inertia of the plate about point O is therefore IO = (Id)O - (Ih)O

= 1.473 kg # m2 - 0.276 kg # m2 = 1.20 kg # m2

Ans.

17.1

ExamplE

17.4

The pendulum in Fig. 17–7 is suspended from the pin at O and consists of two thin rods. Rod OA has a mass of 12 kg, and BC has a mass of 9 kg. Determine the moment of inertia of the pendulum about an axis passing through (a) point O, and (b) the mass center G of the pendulum.

O y–

Solution

2m

part (a). Using the table on the inside back cover, the moment of inertia of rod OA about an axis perpendicular to the page and passing through point O of the rod is IO = 13 ml2. Hence,

This same value can be obtained using IG = theorem.

(IOA ) O =

1 2 12 ml

G A B

1 1 = ml2 = (12 kg)(2 m)2 = 16 kg # m2 3 3

( IOA ) O

417

Mass MoMent of inertia

C 0.75 m

and the parallel-axis

0.75 m

Fig. 17–7

1 1 ml2 + md2 = (12 kg)(2 m)2 + (12 kg)(1 m)2 12 12

= 16 kg # m2 For rod BC we have

(IBC ) O =

17

1 1 ml2 + md2 = (9 kg)(1.5 m)2 + (9 kg)(2 m)2 12 12

= 37.6875 kg # m2

The moment of inertia of the pendulum about O is therefore IO = 16 + 37.6875 = 53.6875 kg # m2 = 53.7 kg # m2

Ans.

part (b). The mass center G will be located relative to point O. Assuming this distance to be y, Fig. 17–7, and using the formula for determining the mass center, we have y =

1(12) + 2(9) ym  = = 1.4286 m m (12) + (9)

The moment of inertia IG may be found in the same manner as IO , which requires successive applications of the parallel-axis theorem to transfer the moments of inertia of rods OA and BC to G. A more direct solution, however, involves using the result for IO , i.e., IO = IG + md2;

53.6875 kg # m2 = IG + (21 kg)(1.4286 m)2 IG = 10.8 kg # m2

Ans.

418

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

prObLeMs 17–1. Determine the moment of inertia Iy for the slender rod. The rod’s density r and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m.

17–3. The solid cylinder has an outer radius R, height h, and is made from a material having a density that varies from its center as r = k + ar 2, where k and a are constants. Determine the mass of the cylinder and its moment of inertia about the z axis.

z

z y

l

R

A

h

x

Prob. 17–1

17

17–2. The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration k x . The density of the material is r = 5 Mg>m3.

Prob. 17–3 *17–4. Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

y y2  50x y 100 mm x

R x

200 mm

Prob. 17–2

Prob. 17–4

17.1 17–5. The hemisphere is formed by rotating the shaded area around the y axis. Determine the moment of inertia Iy and express the result in terms of the total mass m of the hemisphere. The material has a constant density r.

419

Mass MoMent of inertia

17–7. The sphere is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the sphere. The material has a constant density r.

y

y

x2  y2  r2

x2  y2  r2

x

x

Prob. 17–7 17

Prob. 17–5 17–6. The frustum is formed by rotating the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the frustum. The frustum has a constant density r.

*17–8. Determine the mass moment of inertia Iz of the cone formed by revolving the shaded area around the z axis. The density of the material is r. Express the result in terms of the mass m of the cone.

z z

y

h (r r

y)

y  a–b x  b 2b

h

b x

y

z a

Prob. 17–6

x

r

Prob. 17–8

420

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–9. Determine the moment of inertia Iz of the torus. The mass of the torus is m and the density r is constant. Suggestion: Use a shell element.

and

a C C e l e r at i o n

*17–12. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

z

R

O

200 mm

200 mm

a

200 mm

Prob. 17–9 Prob. 17–12 17

17–10. Determine the location y of the center of mass G of the assembly and then calculate the moment of inertia about an axis perpendicular to the page and passing through G. The block has a mass of 3 kg and the semicylinder has a mass of 5 kg. 17–11. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point O. The block has a mass of 3 kg, and the semicylinder has a mass of 5 kg.

17–13. Determine the moment of inertia of the homogeneous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x–y plane and having a thickness dz.

z 400 mm

h z  –– a (x  a) 300 mm

G h –y

200 mm x

b

a

O

Probs. 17–10/11

Prob. 17–13

y

17.1 17–14. The wheel consists of a thin ring having a mass of 10 kg and four spokes made from slender rods and each having a mass of 2 kg. Determine the wheel’s moment of inertia about an axis perpendicular to the page and passing through point A.

Mass MoMent of inertia

421

17–17. The pendulum consists of a 4-kg circular plate and a 2-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

O

500 mm

2m

A

Prob. 17–14 1m

17–15. The assembly is made of the slender rods that have a mass per unit length of 3 kg>m. Determine the mass moment of inertia of the assembly about an axis perpendicular to the page and passing through point O.

17 17–18. Determine the moment of inertia about an axis perpendicular to the page and passing through the pin at O. The thin plate has a hole in its center. Its thickness is 50 mm, and the material has a density r = 50 kg>m3.

O 0.4 m 0.8 m

Prob. 17–17

0.4 m

Prob. 17–15 *17–16. The assembly consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass per unit length of 2 kg>m. Determine the length L of DC so that the center of  mass is at the bearing O. What is the moment of inertia of the assembly about an axis perpendicular to the page and passing through O?

O

150 mm 0.8 m

0.5 m

D

0.2 m

L A

O

1.40 m

1.40 m

B C

Prob. 17–16

Prob. 17–18

422

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–19. The pendulum consists of two slender rods AB and OC which have a mass per unit length of 3 kg>m. The thin circular plate has a mass per unit area of 12 kg>m2. Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. *17–20. The pendulum consists of two slender rods AB and OC which have a mass per unit length of 3 kg>m. The thin circular plate has a mass per unit area of 12 kg>m2. Determine the moment of inertia of the pendulum about an axis perpendicular to the page and passing through the pin at O. 0.4 m

0.4 m

A

B

and

a C C e l e r at i o n

17–22. Determine the moment of inertia of the overhung crank about the x axis. The material is steel having a density of r = 7.85 Mg>m3.

20 mm 30 mm 90 mm 50 mm x

O

180 mm

20 mm

–y

x¿ 30 mm

1.5 m

G

20 mm

30 mm

50 mm

Prob. 17–22

C 0.1 m

17

0.3 m

Probs. 17–19/20 17–21. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. O

17–23. Determine the moment of inertia of the overhung crank about the x  axis. The material is steel having a density of r = 7.85 Mg>m3.

20 mm 30 mm 90 mm

y

50 mm

2m

x

180 mm

20 mm

G

x¿ 30 mm

0.5 m 1m

Prob. 17–21

20 mm

50 mm

30 mm

Prob. 17–23

17.2

17.2

planar KinetiC equations of Motion

423

Planar Kinetic Equations of Motion

In the following analysis we will limit our study of planar kinetics to rigid bodies which, along with their loadings, are considered to be symmetrical with respect to a fixed reference plane.* Since the motion of the body can be viewed within the reference plane, all the forces (and couple moments) acting on the body can then be projected onto the plane. An example of an arbitrary body of this type is shown in Fig. 17–8a. Here the inertial frame of reference x, y, z has its origin coincident with the arbitrary point P in the body. By definition, these axes do not rotate and are either fixed or translate with constant velocity. y F4

F1 M1 V

G M2 W P

F3

x F2

A (a)

Fig. 17–8

Equation of Translational Motion.

The external forces acting on the body in Fig. 17–8a represent the effect of gravitational, electrical, magnetic, or contact forces between adjacent bodies. Since this force system has been considered previously in Sec. 13.3 for the analysis of a system of particles, the resulting Eq. 13–6 can be used here, in which case F = maG This equation is referred to as the translational equation of motion for the mass center of a rigid body. It states that the sum of all the external forces acting on the body is equal to the body’s mass times the acceleration of its mass center G. For motion of the body in the x–y plane, the translational equation of motion may be written in the form of two independent scalar equations, namely, Fx = m(aG)x Fy = m(aG)y *By doing this, the rotational equation of motion reduces to a rather simplified form. The more general case of body shape and loading is considered in Chapter 21.

17

424

Chapter 17

planar KinetiCs

y Fi

x

i

fi y

r P

x

of a

rigid Body: forCe

and

a C C e l e r at i o n

Equation of Rotational Motion. We will now determine the effects caused by the moments of the external force system computed about an axis perpendicular to the plane of motion (the z axis) and passing through point P. As shown on the free-body diagram of the ith particle, Fig. 17–8b, Fi represents the resultant external force acting on the particle, and fi is the resultant of the internal forces caused by interactions with adjacent particles. If the particle has a mass mi and its acceleration is ai , then its kinetic diagram is shown in Fig. 17–8c. Summing moments about point P, we require r * Fi + r * fi = r * mi ai

Particle free-body diagram

or

(b)

(MP)i = r * mi ai

=

The moments about P can also be expressed in terms of the acceleration of point P, Fig. 17–8d. If the body has an angular acceleration A and angular velocity V, then using Eq. 16–18 we have

y

(MP)i = mi r * (aP + A * r - v2r) miai

x

i

The last term is zero, since r * r = 0. Expressing the vectors with Cartesian components and carrying out the cross-product operations yields

17 y

r P

(MP)i k = mi 5 (xi + yj) * [(aP)x i + (aP)y j]

x

+ (xi + yj) * [ak * (xi + yj)] 6

Particle kinetic diagram

(MP)i k = mi[-y(aP)x + x(aP)y + ax2 + ay2]k

(c)

a(MP)i = mi[-y(aP)x + x(aP)y + ar2] Letting mi S dm and integrating with respect to the entire mass m of the body, we obtain the resultant moment equation

y

_ x

aG

aMP = - a

Lm

y dmb(aP)x + a

x dmb (aP)y + a r2dmb a Lm Lm

(d)

Here MP represents only the moment of the external forces acting on the body about point P. The resultant moment of the internal forces is zero, since for the entire body these forces occur in equal and opposite collinear pairs and thus the moment of each pair of forces about P cancels. The integrals in the first and second terms on the right are used to locate the body’s center of mass G with respect to P, since ym = 1 y dm and xm = 1 x dm, Fig. 17–8d. Also, the last integral represents the body’s moment of inertia about the z axis, i.e., IP = 1 r2dm. Thus,

Fig. 17–8 (cont.)

aMP = -ym(aP)x + xm(aP)y + IPa

G

_ r P

= mi[r * aP + r * (A * r) - v2(r * r)]

V _ y

aP

x

A

(17–6)

17.2

It is possible to reduce this equation to a simpler form if point P coincides with the mass center G for the body. If this is the case, then x = y = 0, and therefore* MG = IGa

y F4

F1 M1

(17–7)

This rotational equation of motion states that the sum of the moments of all the external forces about the body’s mass center G is equal to the product of the moment of inertia of the body about an axis passing through G and the body’s angular acceleration. Equation 17–6 can also be rewritten in terms of the x and y components of aG and the body’s moment of inertia IG . If point G is located at (x, y), Fig. 17–8d, then by the parallel-axis theorem, IP = IG + m(x2 + y2). Substituting into Eq. 17–6 and rearranging terms, we get aMP = ym[-(aP)x + ya] + xm[(aP)y + xa] + IGa

G M2 W P

Free-body diagram

F2

(e)

(17–8) y m(aG)y

aG = aP + A * r - v2r (aG)x i + (aG)y j = (aP)x i + (aP)y j + ak * (x i + y j) - v2(x i + y j)

_ x

Carrying out the cross product and equating the respective i and j components yields the two scalar equations (aG)x = (aP)x - ya - xv2 (aG)y = (aP)y + xa - yv2

IG A G

_ y

m(aG)x

P

From these equations, [-(aP)x + ya] = [-(aG)x - xv2] and [(aP)y + xa] = [(aG)y + yv2]. Substituting these results into Eq. 17–8 and simplifying gives (17–9)

This important result indicates that when moments of the external forces shown on the free-body diagram are summed about point P, Fig. 17–8e, they are equivalent to the sum of the “kinetic moments” of the components of maG about P plus the “kinetic moment” of IG A, Fig. 17–8f. In other words, when the “kinetic moments,” (mk)P , are computed, Fig.  17–8f, the vectors m(aG)x and m(aG)y are treated as sliding vectors; that is, they can act at any point along their line of action. In a similar manner, IG A can be treated as a free vector and can therefore act at any point. It is important to keep in mind, however, that maG and IG A are not the same as a force or a couple moment. Instead, they are caused by the external effects of forces and couple moments acting on the body. With this in mind we can therefore write Eq. 17–9 in a more general form as MP = (mk)P

x

F3

From the kinematic diagram of Fig. 17–8d, aP can be expressed in terms of aG as

aMP = -ym(aG)x + xm(aG)y + IGa

425

planar KinetiC equations of Motion

(17–10)

*It also reduces to this same simple form MP = IPa if point P is a fixed point (see Eq. 17–16) or the acceleration of point P is directed along the line PG.

17 x

Kinetic diagram (f)

Fig. 17–8 (cont.)

426

Chapter 17

planar KinetiCs

of a

and

a C C e l e r at i o n

General Application of the Equations of Motion.

y F4

F1 M1

To summarize this analysis, three independent scalar equations can be written to describe the general plane motion of a symmetrical rigid body. Fx = m ( aG ) x

G M2

Fy = m ( aG ) y

W P

MG = IGa

x

Free-body diagram

F3

rigid Body: forCe

F2

or

(17–11)

When applying these equations, one should always draw a free-body diagram, Fig. 17–8e, in order to account for the terms involved in Fx , Fy , MG , or MP . In some problems it may also be helpful to draw the kinetic diagram for the body, Fig. 17–8f. This diagram graphically accounts for the terms m(aG)x , m(aG)y , and IG A. It is especially convenient when used to determine the components of maG and the moment of these components in (mk)P .*

(e)

y m(aG)y _ x

M P = (k )P

IG A G

17

m(aG)x

_ y

P

x

17.3

Equations of Motion: Translation

When the rigid body in Fig. 17–9a undergoes a translation, all the particles of the body have the same acceleration. Furthermore, A = 0, in which case the rotational equation of motion applied at point G reduces to a simplified form, namely, MG = 0. Application of this and the force equations of motion will now be discussed for each of the two types of translation.

Kinetic diagram (f)

Fig. 17–8 (cont.)

Rectilinear Translation. When a body is subjected to rectilinear

translation, all the particles of the body (slab) travel along parallel straightline paths. The free-body and kinetic diagrams are shown in Fig.  17–9b. Since IG A = 0, only maG is shown on the kinetic diagram. Hence, the equations of motion which apply in this case become

F1 F4

M1 G F2

M2

Fx = m(aG)x Fy = m(aG)y MG = 0

(17–12)

F3 (a)

Fig. 17–9

*For this reason, the kinetic diagram will be used in the solution of an example problem whenever M P = (k )P is applied.

F1 A M2

M1

G

d

=

F4

R Tr ect an ilin sla ea tio r n

17.3

F2

427

equations of Motion: translation

maG

A G

W F3 (b)

It is also possible to sum moments about other points on or off the body, in which case the moment of maG must be taken into account. For example, if point A is chosen, which lies at a perpendicular distance d from the line of action of maG , the following moment equation applies: a + M A = (k )A ;

MA = (maG)d

Here the sum of moments of the external forces and couple moments about A (MA , free-body diagram) equals the moment of maG about A ((k )A, kinetic diagram).

17 t

F1

Curvilinear Translation. When a rigid body is subjected to curvilinear

translation, all the particles of the body have the same accelerations as they travel along curved paths as noted in Sec.16.1. For analysis, it is often convenient to use an inertial coordinate system having an origin which coincides with the body’s mass center at the instant considered, and axes which are oriented in the normal and tangential directions to the path of motion, Fig. 17–9c. The three scalar equations of motion are then

F4

M2 B

ilinear Curv lation n a r T s M1 F2

G W

n F3

=

Fn = m(aG)n Ft = m(aG)t MG = 0

(17–13)

t

m(aG)t

If moments are summed about the arbitrary point B, Fig. 17–9c, then it is necessary to account for the moments, (k )B , of the two components m(aG)n and m(aG)t about this point. From the kinetic diagram, h and e represent the perpendicular distances (or “moment arms”) from B to the lines of action of the components. The required moment equation therefore becomes a + MB = (mk)B;

MB = e[m(aG)t] - h[m(aG)n]

h B

G m(aG)n

e n (c)

Fig. 17–9

428

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

procedure for analysis Kinetic problems involving rigid-body translation can be solved using the following procedure. The free-body and kinetic diagrams for this boat and trailer are drawn first in order to apply the equations of motion. Here the forces on the free-body diagram cause the effect shown on the kinetic diagram. If moments are summed about the mass center, G, then MG = 0. However, if moments are summed about point B then c + MB = maG(d).

Free-Body Diagram. • Establish the x, y or n, t inertial coordinate system and draw the free-body diagram in order to account for all the external forces and couple moments that act on the body.

• The direction and sense of the acceleration of the body’s mass center aG should be established.

• Identify the unknowns in the problem. • If it is decided that the rotational equation of motion

MP = (mk)P is to be used in the solution, then consider drawing the kinetic diagram, since it graphically accounts for the components m(aG)x , m(aG)y or m(aG)t , m(aG)n and is therefore convenient for “visualizing” the terms needed in the moment sum (mk)P .

W

17

G A

B NB

T NA

= G B

d

maG

Equations of motion. • Apply the three equations of motion in accordance with the established sign convention. • To simplify the analysis, the moment equation MG = 0 can be replaced by the more general equation MP = (mk)P , where point P is usually located at the intersection of the lines of action of as many unknown forces as possible. • If the body is in contact with a rough surface and slipping occurs, use the friction equation F = mkN. Remember, F always acts on the body so as to oppose the motion of the body relative to the surface it contacts. Kinematics. • Use kinematics to determine the velocity and position of the body. • For rectilinear translation with variable acceleration aG = dvG >dt aGdsG = vGdvG

• For rectilinear translation with constant acceleration vG = (vG)0 + aGt v2G = (vG)20 + 2aG[sG - (sG)0] sG = (sG)0 + (vG)0t + 12 aGt2

• For curvilinear translation (aG)n = v2G >r

(aG)t = dvG >dt

(aG)t dsG = vG dvG

17.3

ExamplE

17.5

The car shown in Fig. 17–10a has a mass of 2 Mg and a center of mass at G. Determine the acceleration if the rear “driving” wheels are always slipping, whereas the front wheels are free to rotate. Neglect the mass of the wheels. The coefficient of kinetic friction between the wheels and the road is mk = 0.25.

G A

equations of Motion.

0.3 m

1.25 m 0.75 m

B

(a)

Solution i Free-body Diagram. As shown in Fig. 17–10b, the rear-wheel frictional force FB pushes the car forward, and since slipping occurs, FB = 0.25NB . The frictional forces acting on the front wheels are zero, since these wheels have negligible mass.* There are three unknowns in the problem, NA , NB , and aG . Here we will sum moments about the mass center. The car (point G) accelerates to the left, i.e., in the negative x direction, Fig. 17–10b.

y aG x 2000 (9.81) N

0.3 m

G

FB  0.25 NB

A

+ Fx = m(aG)x; S

-0.25NB = -(2000 kg)aG

(1)

+ c Fy = m(aG)y;

NA + NB - 2000(9.81) N = 0

(2)

a + MG = 0;

429

equations of Motion: translation

NA

NB 1.25 m

0.75 m (b)

-NA(1.25 m) - 0.25NB(0.3 m) + NB(0.75 m) = 0 (3)

17

Solving, aG = 1.59 m>s2 d

Ans.

2000 (9.81) N

NA = 6.88 kN NB = 12.7 kN

NB(2 m) - [2000(9.81) N](1.25 m) = (2000 kg)aG(0.3 m)

Solving this and Eq. 1 for aG leads to a simpler solution than that obtained from Eqs. 1 to 3.

*With negligible wheel mass, Ia = 0 and the frictional force at A required to turn the wheel is zero. If the wheels’ mass were included, then the solution would be more involved, since a general-plane-motion analysis of the wheels would have to be considered (see Sec. 17.5).

FB  0.25 NB

A NB

NA

1.25 m

0.75 m

=

Solution ii Free-body and Kinetic Diagrams. If the “moment” equation is applied about point A, then the unknown NA will be eliminated from the equation. To “visualize” the moment of maG about A, we will include the kinetic diagram as part of the analysis, Fig. 17–10c. equation of Motion. a + MA = (mk)A;

0.3 m

G

2000 aG

G 0.3 m

A (c)

Fig. 17–10

430

Chapter 17

ExamplE

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

17.6 The motorcycle shown in Fig. 17–11a has a mass of 125 kg and a center of mass at G1 , while the rider has a mass of 75 kg and a center of mass at G2 . Determine the minimum coefficient of static friction between the wheels and the pavement in order for the rider to do a “wheely,” i.e., lift the front wheel off the ground as shown in the photo. What acceleration is necessary to do this? Neglect the mass of the wheels and assume that the front wheel is free to roll.

735.75 N 1226.25 N

G2

0.3 m

G1

0.6 m FB

B 0.4 m 0.4 m NB

B

0.7 m A NA  0

=

17

75 kg aG 125 kg aG 0.3 m 0.6 m B (b)

Fig. 17–11

0.4 m 0.4 m (a)

0.7 m

A

Solution Free-body and Kinetic Diagrams. In this problem we will consider both the motorcycle and the rider as a single system. It is possible first to determine the location of the center of mass for this “system” by using x m> m and y =   y m> m. Here, however, we will the equations x =   consider the weight and mass of the motorcycle and rider separately as shown on the free-body and kinetic diagrams, Fig. 17–11b. Both of these parts move with the same acceleration. We have assumed that the front wheel is about to leave the ground, so that the normal reaction NA  0. The three unknowns in the problem are NB , FB , and aG . equations of Motion. + Fx = m(aG)x; S

FB = (75 kg + 125 kg)aG

+ c Fy = m(aG)y;

NB - 735.75 N - 1226.25 N = 0

(1)

a + MB = (mk)B; -(735.75 N)(0.4 m) - (1226.25 N)(0.8 m) = -(75 kg aG)(0.9 m) - (125 kg aG)(0.6 m) Solving,

aG = 8.95 m>s2 S

(2) Ans.

NB = 1962 N FB = 1790 N Thus the minimum coefficient of static friction is (ms)min =

FB 1790 N = = 0.912 NB 1962 N

Ans.

17.3

ExamplE

431

equations of Motion: translation

17.7

The 100-kg beam BD shown in Fig. 17–12a is supported by two rods having negligible mass. Determine the force developed in each rod if at the instant u = 30, v = 6 rad>s. Solution Free-body and Kinetic Diagrams. The beam moves with curvilinear translation since all points on the beam move along circular paths, each path having the same radius of 0.5 m, but different centers of curvature. Using normal and tangential coordinates, the free-body and kinetic diagrams for the beam are shown in Fig. 17–12b. Because of the translation, G has the same motion as the pin at B, which is connected to both the rod and the beam. Note that the tangential component of acceleration acts downward to the left due to the clockwise direction of A, Fig. 17–12c. Furthermore, the normal component of acceleration is always directed toward the center of curvature (toward point A for rod AB). Since the angular velocity of AB is 6 rad>s when u = 30, then

A

C

V u  30

0.5 m

G B

0.4 m

0.4 m

D

(a)

(aG)n = v2r = (6 rad>s)2(0.5 m) = 18 m>s2 The three unknowns are TB , TD , and (aG)t . TB

30

TD

30

100 kg(aG)n A

30

G 0.4 m

17

0.4 m

=

A v  6 rad/s an

0.5 m 100 kg(aG)t

B

at

981 N

(c) (b)

Fig. 17–12

equations of Motion. + aFn = m(aG)n; TB + TD - 981 cos 30 N = 100 kg(18 m>s2)

(1)

+ bFt = m(aG)t; 981 sin 30 = 100 kg(aG)t

(2)

a + MG = 0; -(TB cos 30)(0.4 m) + (TD cos 30)(0.4 m) = 0

(3)

Simultaneous solution of these three equations gives TB = TD = 1.32 kN

Ans.

(aG)t = 4.905 m>s2 NoTE: It is also possible to apply the equations of motion along horizontal and vertical x, y axes, but the solution becomes more involved.

432

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

preLiMinary prObLeMs P17–1. Draw the free-body and kinetic diagrams of the object AB.

100 kg G

0.5 m

5 4

0.5 m A mk  0.2

3m

3

B

2m

2m

A

100 N

1m

2m G

B

100 kg

1m

4 rad/s

(a)

(d) 100 kg

100 kg

G

0.5 m

0.5 m

A

A

2m

17

G B 2m

2m 3m

3 rad/s

B

60

1.5 m (e)

(b)

100 kg

B

500 N

2m

G

A 2m G

0.5

100 kg

30

A 5

B m  0.2 k

m

1.5 m

1m

3

4

(c)

(f)

Prob. P17–1

17.3

equations of Motion: translation

433

P17–2. Draw the free-body and kinetic diagrams of the 100-kg object.

20 N  m O

100 N 2 rad/s 3m 2m (a) v  4 rad/s O (d)

4 rad/s O 45

17 O

2m

3m

v  3 rad/s

60 N 45

(e)

(b)

5m O 1m

2 rad/s k  6 N/m

2m

O

4m

2 rad/s

3m

30 N  m

Unstretched length of spring is 1 m. (f)

(c)

Prob. P17–2

434

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

a C C e l e r at i o n

and

FunDaMentaL prObLeMs F17–1. The cart and its load have a total mass of 100 kg. Determine the acceleration of the cart and the normal reactions on the pair of wheels at A and B. Neglect the mass of the wheels. 100 N 3

5 4

F17–4. Determine the maximum acceleration of the truck without causing the assembly to move relative to the truck. Also what is the corresponding normal reaction on legs A and B? The [email protected] table has a mass center at G and the coefficient of static friction between the legs of the table and the bed of the truck is ms = 0.2. 0.6 m 0.9 m G

1.2 m G

0.75 m

0.5 m B

a A

B

A

Prob. F17–1

Prob. F17–4

F17–2. If the 80-kg cabinet is allowed to roll down the inclined plane, determine the acceleration of the cabinet and the normal reactions on the pair of rollers at A and B that have negligible mass.

F17–5. At the instant shown both rods of negligible mass swing with a counterclockwise angular velocity of v = 5 rad>s, while the [email protected] bar is subjected to the [email protected] horizontal force. Determine the tension developed in the rods and the angular acceleration of the rods at this instant.

0.6 m

0.3 m 0.4 m

17

A G

C

v  5 rad/s 1.5 m

1.5 m

15

100 N

A B

0.5 m

B

0.5 m

Prob. F17–2 F17–3. The [email protected] link AB is pinned to a moving frame at A and held in a vertical position by means of a string BC which can support a maximum tension of 50 N. Determine the maximum acceleration of the frame without breaking the string. What are the corresponding components of reaction at the pin A?

D G 1m

1m

Prob. F17–5 F17–6. At the instant shown, link CD rotates with an angular velocity of v = 6 rad>s. If it is subjected to a couple moment M = 450 N # m, determine the force developed in link AB, the horizontal and vertical component of reaction on pin D, and the angular acceleration of link CD at this instant. The block has a mass of 50 kg and center of mass at G. Neglect the mass of links AB and CD. 0.1 m

A

0.6 m

a A

B

0.9 m 0.4 m

C

G

1.2 m

0.4 m B 0.9 m

D

Prob. F17–3

v  6 rad/s

C

M  450 Nm

Prob. F17–6

17.3

435

equations of Motion: translation

prObLeMs *17–24. The jet aircraft has a total mass of 22 Mg and a center of mass at G. Initially at take-off the engines provide a thrust 2T = 4 kN and T = 1.5 kN. Determine the acceleration of the plane and the normal reactions on the nose wheel at A and each of the two wing wheels located at B. Neglect the mass of the wheels and, due to low velocity, neglect any lift caused by the wings.

T¿ 2.5 m

G

2T

1.2 m

2.3 m

B 3m

6m

A

17–27. The assembly has a mass of 8 Mg and is hoisted using the boom and pulley system. If the winch at B draws in the cable with an acceleration of 2 m>s2, determine the compressive force in the hydraulic cylinder needed to support the boom. The boom has a mass of 2 Mg and mass center at G. *17–28. The assembly has a mass of 4 Mg and is hoisted using the winch at B. Determine the greatest acceleration of the assembly so that the compressive force in the hydraulic cylinder supporting the boom does not exceed 180 kN. What is the tension in the supporting cable? The boom has a mass of 2 Mg and mass center at G.

Prob. 17–24

6m

17–25. The 4-Mg uniform canister contains nuclear waste material encased in concrete. If the mass of the spreader beam BD is 50 kg, determine the force in each of the links AB, CD, EF, and GH when the system is lifted with an acceleration of a = 2 m>s2 for a short period of time. 17–26. The 4-Mg uniform canister contains nuclear waste material encased in concrete. If the mass of the spreader beam BD is 50 kg, determine the largest vertical acceleration a of the system so that each of the links AB and CD are not subjected to a force greater than 30 kN and links EF and GH are not subjected to a force greater than 34 kN.

2m

G

17 C

4m B 1m

60$ A

D

2m

a

Probs. 17–27/28 F

H 30

30 B

E 0.3 m

0.4 m

G

D

0.3 m

17–29. Determine the shortest time possible for the rearwheel drive, 2-Mg truck to achieve a speed of 16 m>s with a constant acceleration starting from rest. The coefficient of static friction between the wheels and the road surface is ms = 0.8. The front wheels are free to roll. Neglect the mass of the wheels.

C

A

G 0.75 m A

B 2m

Probs. 17–25/26

Prob. 17–29

1.5 m

436

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–30. The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to slip or tip on the cart. The coefficient of static friction between the crate and cart is ms = 0.2.

and

a C C e l e r at i o n

*17–32. The pipe has a mass of 460 kg and is held in place on the truck bed using the two boards A and B. Determine the greatest acceleration of the truck so that the pipe begins to lose contact at A and the bed of the truck and starts to pivot about B. Assume board B will not slip on the bed of the truck, and the pipe is smooth. Also, what force does board B exert on the pipe during the acceleration?

0.5 m 0.4 m

0.5 m

P A

B 1m

0.3 m

Prob. 17–32

17

Prob. 17–30

17–31. The 150-kg uniform crate rests on the 10-kg cart. Determine the maximum force P that can be applied to the handle without causing the crate to tip on the cart. Slipping does not occur.

17–33. The uniform girder AB has a mass of 8 Mg. Determine the internal axial, shear, and bending-moment loadings at the center of the girder if a crane gives it an upward acceleration of 3 m>s2.

3 m/s2 C

0.5 m

P 1m

A

Prob. 17–31

60

4m

Prob. 17–33

60

B

17–34. The mountain bike has a mass of 40 kg with center of mass at point G1, while the rider has a mass of 60 kg with center of mass at point G2. Determine the maximum deceleration when the brake is applied to the front wheel, without causing the rear wheel B to leave the road. Assume that the front wheel does not slip. Neglect the mass of all the wheels. 17–35. The mountain bike has a mass of 40 kg with center of mass at point G1, while the rider has a mass of 60 kg with center of mass at point G2. When the brake is applied to the front wheel, it causes the bike to decelerate at a constant rate of 3 m>s2. Determine the normal reaction the road exerts on the front and rear wheels. Assume that the rear wheel is free to roll. Neglect the mass of all the wheels.

437

equations of Motion: translation

17.3

17–37. A force of P = 300 N is applied to the 60-kg cart. Determine the reactions at both the wheels at A and both the wheels at B. Also, what is the acceleration of the cart? The mass center of the cart is at G.

P 30 G 0.4 m 0.3 m B

A

G2

0.3 m

1.25 m

G1

0.2 m

0.08 m

Prob. 17–37 17

0.4 m A

B 0.4 m

17–38. Determine the largest force P that can be applied to the 60-kg cart, without causing one of the wheel reactions, either at A or at B, to be zero. Also, what is the acceleration of the cart? The mass center of the cart is at G.

0.4 m 0.2 m

Probs. 17–34/35 *17–36. The trailer with its load has a mass of 150-kg and a center of mass at G. If it is subjected to a horizontal force of P = 600 N, determine the trailer’s acceleration and the normal force on the pair of wheels at A and at B. The wheels are free to roll and have negligible mass.

P 30 G 0.4 m

G

0.3 m P  600 N

1.25 m 0.25 m

0.25 m

B

A

0.5 m

B

A 0.3 m

0.75 m

1.25 m

Prob. 17–36

0.2 m

0.08 m

Prob. 17–38

438

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–39. If the cart’s mass is 30 kg and it is subjected to a horizontal force of P = 90 N, determine the tension in cord AB and the horizontal and vertical components of reaction on end C of the uniform 15-kg rod BC. *17–40. If the cart’s mass is 30 kg, determine the horizontal force P that should be applied to the cart so that the cord AB just becomes slack. The uniform rod BC has a mass of 15 kg.

30

C

a C C e l e r at i o n

17–42. The dragster has a mass of 1500 kg and a center of mass at G. If the coefficient of kinetic friction between the rear wheels and the pavement is mk = 0.6, determine if it is possible for the driver to lift the front wheels, A, off the ground while the rear drive wheels are slipping. Neglect the mass of the wheels and assume that the front wheels are free to roll.

17–43. The dragster has a mass of 1500 kg and a center of mass at G. If no slipping occurs, determine the frictional force FB which must be developed at each of the rear drive wheels B in order to create an acceleration of a = 6 m>s2. What are the normal reactions of each wheel on the ground? Neglect the mass of the wheels and assume that the front wheels are free to roll.

A

1m

and

B

30

P 0.25 m

17

0.3 m

Probs. 17–39/40

G B

1m

17–41. The uniform bar of mass m is pin connected to the collar, which slides along the smooth horizontal rod. If the collar is given a constant acceleration of a, determine the bar's inclination angle u. Neglect the collar's mass.

A

L

a

2.5 m

Probs. 17–42/43

*17–44. The pipe has a length of 3 m and a mass of 500 kg. It is attached to the back of the truck using a 0.6-m-long chain AB. If the coefficient of kinetic friction at C is mk = 0.4, determine the acceleration of the truck if the angle u = 10 with the road as shown.

u

B 3m

A θ = 10°

1m

C

Prob. 17–41

A

Prob. 17–44

17–45. The lift truck has a mass of 70 kg and mass center at G. If it lifts the 120-kg spool with an acceleration of 3 m>s2, determine the reactions of each of the four wheels. The loading is symmetric. Neglect the mass of the movable arm CD. 17–46. The lift truck has a mass of 70 kg and mass center at G. Determine the largest upward acceleration of the 120-kg spool so that no reaction of the wheels on the ground exceeds 600 N.

439

equations of Motion: translation

17.3

*17–48. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the tension in the cable when the truck begins to accelerate at 5 m>s2. Also, what are the horizontal and vertical components of reaction at the hinge C? 17–49. The drop gate at the end of the trailer has a mass of 1.25 Mg and mass center at G. If it is supported by the cable AB and hinge at C, determine the maximum deceleration of the truck so that the gate does not begin to rotate forward. What are the horizontal and vertical components of reaction at the hinge C?

0.7 m

B

A G

C

D

C

30

G

1m

1.5 m 45

0.4 m A

Probs. 17–48/49

B 0.75 m

17

0.5 m

Probs. 17–45/46 17–47. The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration? The coefficient of static friction between the crate and cart is ms = 0.5.

17–50. The bar has a weight per length w and is supported by the smooth collar. If it is released from rest, determine the internal normal force, shear force, and bending moment in the bar as a function of x.

0.6 m 30

F

1m

x

15

Prob. 17–47

Prob. 17–50

440

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–51. The pipe has a mass of 800 kg and is being towed behind the truck. If the acceleration of the truck is at = 0.5 m>s2, determine the angle u and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

B

and

a C C e l e r at i o n

17–53. The 100-kg uniform crate C rests on the elevator floor where the coefficient of static friction is ms = 0.4. Determine the largest initial angular acceleration a, starting from rest at u = 90, without causing the crate to slip. No tipping occurs.

0.6 m

at

1.2 m

C A G 0.4 m

45

E

1.5 m

B 1.5 m

u D

u

a

u

A

C

Prob. 17–53

Prob. 17–51 17

*17–52. The pipe has a mass of 800 kg and is being towed behind a truck. If the angle u = 30, determine the acceleration of the truck and the tension in the cable. The coefficient of kinetic friction between the pipe and the ground is mk = 0.1.

17–54. The two uniform 4-kg bars DC and EF are fixed (welded) together at E. Determine the normal force N E , shear force V E , and moment M E , which DC exerts on EF at E if at the instant u = 60. BC has an angular velocity v = 2 rad>s and an angular acceleration a = 4 rad>s2 as shown.

F

B

1.5 m

at E

D

A G 0.4 m

C

45 2m

2m

u u  60 A

a  4 rad/s2 B

v  2 rad/s

C

Prob. 17–52

Prob. 17–54

17.4

441

equations of Motion: rotation aBout a fixed axis

17–55. The arched pipe has a mass of 80 kg and rests on the surface of the platform for which the coefficient of static friction is ms = 0.3. Determine the greatest angular acceleration a of the platform, starting from rest when u = 45, without causing the pipe to slip on the platform.

*17–56. Determine the force developed in the links and the acceleration of the bar’s mass center immediately after the cord fails. Neglect the mass of links AB and CD. The uniform bar has a mass of 20 kg.

A 500 mm

45 G

A 1m

θ

C

B

0.4 m

200 mm

1m

P

D

B

ω, α

50 N

0.6 m

Prob. 17–55

Prob. 17–56 17

17.4

Equations of Motion: Rotation about a Fixed Axis

Consider the rigid body (or slab) shown in Fig. 17–13a, which is constrained to rotate in the vertical plane about a fixed axis perpendicular to the page and passing through the pin at O. The angular velocity and angular acceleration are caused by the external force and couple moment system acting on the body. Because the body’s center of mass G moves around a circular path, the acceleration of this point is best represented by its tangential and normal components. The tangential component of acceleration has a magnitude of (aG)t = arG and must act in a direction which is consistent with the body’s angular acceleration A. The magnitude of the normal component of acceleration is (aG)n = v2rG . This component is always directed from point G to O, regardless of the rotational sense of V.

F3 A V

M1 M2

F2 (aG)t

G rG O

(aG)n

F1 (a)

Fig. 17–13

F4

442

Chapter 17

planar KinetiCs

F3 A V

M1 M2

F2 (aG)t

F4

G rG O

of a

rigid Body: forCe

and

a C C e l e r at i o n

The free-body and kinetic diagrams for the body are shown in Fig. 17–13b. The two components m(aG)t and m(aG)n , shown on the kinetic diagram, are associated with the tangential and normal components of acceleration of the body’s mass center. The IG A vector acts in the same direction as A and has a magnitude of IGa, where IG is the body’s moment of inertia calculated about an axis which is perpendicular to the page and passes through G. From the derivation given in Sec. 17.2, the equations of motion which apply to the body can be written in the form

(aG)n

Fn = m(aG)n = mv2rG Ft = m(aG)t = marG MG = IGa

F1 (a)

(17–14)

The moment equation can be replaced by a moment summation about any arbitrary point P on or off the body provided one accounts for the moments (mk)P produced by IG A, m(aG)t , and m(aG)n about the point.

Moment Equation About Point O. Often it is convenient to

sum moments about the pin at O in order to eliminate the unknown force FO . From the kinetic diagram, Fig. 17–13b, this requires

17

F3

a + MO = (mk)O;

M1 M2

F2 G

F4 W

O FO

m(aG)t

IGA G

(17–15)

Note that the moment of m(aG)n is not included here since the line of action of this vector passes through O. Substituting (aG)t = rGa, we may rewrite the above equation as a + MO = (IG + mr2G)a. From the parallelaxis theorem, IO = IG + md2, and therefore the term in parentheses represents the moment of inertia of the body about the fixed axis of rotation passing through O.* Consequently, we can write the three equations of motion for the body as Fn = m(aG)n = mv2rG Ft = m(aG)t = marG MO = IOa

=

F1

MO = rGm(aG)t + IGa

(17–16)

When using these equations, remember that ;IOa< accounts for the “moment” of both m(aG)t and IG A about point O, Fig. 17–13b. In other words, MO = (mk)O = IOa, as indicated by Eqs. 17–15 and 17–16.

m(aG)n O

rG

(b)

Fig. 17–13 (cont.)

*The result MO = IOa can also be obtained directly from Eq. 17–6 by selecting point P to coincide with O, realizing that (aP)x = (aP)y = 0.

17.4

equations of Motion: rotation aBout a fixed axis

443

procedure for analysis Kinetic problems which involve the rotation of a body about a fixed axis can be solved using the following procedure. Free-Body Diagram. • Establish the inertial n, t coordinate system and specify the direction and sense of the accelerations (aG)n and (aG)t and the angular acceleration A of the body. Recall that (aG)t must act in a direction which is in accordance with the rotational sense of A, whereas (aG)n always acts toward the axis of rotation, point O.

• Draw the free-body diagram to account for all the external forces and couple moments that act on the body.

• Determine the moment of inertia IG or IO . • Identify the unknowns in the problem. • If it is decided that the rotational equation of motion

Equations of motion. • Apply the three equations of motion in accordance with the established sign convention.

G M Ox

W

O Oy

m(aG)t IGA

• If moments are summed about the body’s mass center, G, then MG = IGa, since (maG)t and (maG)n create no moment about G.

• If moments are summed about the pin support O on the axis of

rotation, then (maG)n creates no moment about O, and it can be shown that MO = IOa.

Kinematics. • Use kinematics if a complete solution cannot be obtained strictly from the equations of motion.

• If the angular acceleration is variable, use dv du a du = v dv v = dt dt If the angular acceleration is constant, use a =



v = v0 + act u = u0 + v0t + 12 act2 v2 = v20 + 2ac(u - u0)

17

=

MP = (mk)P is to be used, i.e., P is a point other than G or O, then consider drawing the kinetic diagram in order to help “visualize” the “moments” developed by the components m(aG)n , m(aG)t , and IG A when writing the terms for the moment sum (mk)P .

T

m(aG)n O

G

d

The crank on the oil-pumping rig undergoes rotation about a fixed axis which is caused by a driving torque M of the motor. The loadings shown on the free-body diagram cause the effects shown on the kinetic diagram. If moments are summed about the mass center,  G, then MG = IGa. However, if moments are summed about point O, noting that (aG)t = ad, then a + M O = IGa + m(aG)t d + m(aG)n(0) = (IG + md2)a = IOa.

444

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

17.8

ExamplE

The unbalanced 25-kg flywheel shown in Fig. 17–14a has a radius of gyration of kG = 0.18 m about an axis passing through its mass center G. If it is released from rest, determine the horizontal and vertical components of reaction at the pin O.

0.15 m O

G

Solution Free-body and Kinetic Diagrams. Since G moves in a circular path, it will have both normal and tangential components of acceleration. Also, since a, which is caused by the flywheel’s weight, acts clockwise, the tangential component of acceleration must act downward. Why? Since v = 0, only m(aG)t = marG and IGa are shown on the kinetic diagram in Fig. 17–14b. Here, the moment of inertia about G is

(a)

IG = mk2G = (25 kg)(0.18 m)2 = 0.81 kg # m2

The three unknowns are On , Ot , and a. equations of Motion. + Fn = mv2rG; d + T Ft = marG;

n

17 0.15 m On

O

Ot

t

Ans.

Moments can also be summed about point O in order to eliminate On and Ot and thereby obtain a direct solution for A, Fig. 17–14b. This can be done in one of two ways.

=

c + MO = (mk)O;

[25(9.81)N] (0.15m) = (0.81 kg # m2)a +

3 (25 kg)a(0.15 m) 4 (0.15 m)

245.25 N(0.15 m) = 1.3725a

G

Fig. 17–14

(1)

Solving,

IGa

(b)

Ans.

Ot(0.15 m) = (0.81 kg # m2)a a = 26.8 rad>s2 Ot = 144.7 N

25(9.81)N

marG

-Ot + 25(9.81)N = (25 kg)(a)(0.15 m)

c + MG = IGa;

G

rG O

On = 0

(2)

If MO = IOa is applied, then by the parallel-axis theorem the moment of inertia of the flywheel about O is IO = IG + mr2G = 0.81 + (25)(0.15)2 = 1.3725 kg # m2

Hence, c + MO = IOa;

(245.25 N)(0.15 m) = (1.3725 kg # m2)a

which is the same as Eq. 2. Solving for a and substituting into Eq. 1 yields the answer for Ot obtained previously.

17.4

ExamplE

445

equations of Motion: rotation aBout a fixed axis

17.9

At the instant shown in Fig. 17–15a, the 20-kg slender rod has an angular velocity of v = 5 rad>s. Determine the angular acceleration and the horizontal and vertical components of reaction of the pin on the rod at this instant. O

60 N  m v  5 rad/s 3m (a)

Solution Free-body and Kinetic Diagrams. Fig. 17–15b. As shown on the kinetic diagram, point G moves around a circular path and so it has two components of acceleration. It is important that the tangential component at = arG act downward since it must be in accordance with the rotational sense of A. The three unknowns are On , Ot , and a.

60 N  m

On O Ot

1.5 m 20(9.81) N

=

equation of Motion. + Fn = mv2rG; d

On = (20 kg)(5 rad>s)2(1.5 m)

+ T Ft = marG;

-Ot + 20(9.81)N = (20 kg)(a)(1.5 m)

c + MG = IGa;

G

Ot(1.5 m) + 60 N # m =

Solving

17 mv2rG

3 121 (20 kg)(3 m)2 4 a

On = 750 N Ot = 19.05 N a = 5.90 rad>s2

Ans.

A more direct solution to this problem would be to sum moments about point O to eliminate On and Ot and obtain a direct solution for a. Here, c + MO = (mk)O; 60 N # m + 20(9.81) N(1.5 m) =

3 121 (20 kg)(3 m)2 4 a

+ [20 kg(a)(1.5 m)](1.5 m)

a = 5.90 rad>s2

Ans.

Also, since IO = 13 ml2 for a slender rod, we can apply c + MO = IOa; 60 N # m + 20(9.81) N(1.5 m) = a = 5.90 rad>s2

3 13(20 kg)(3 m)2 4 a

Ans.

NoTE: By comparison, the last equation provides the simplest solution

for a and does not require use of the kinetic diagram.

IGA

O

G

rG marG (b)

Fig. 17–15

446

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

17.10

ExamplE

The drum shown in Fig. 17–16a has a mass of 60 kg and a radius of gyration kO = 0.25 m. A cord of negligible mass is wrapped around the periphery of the drum and attached to a block having a mass of 20 kg. If the block is released, determine the drum’s angular acceleration.

0.4 m O

A

Solution i Free-body Diagram. Here we will consider the drum and block separately, Fig. 17–16b. Assuming the block accelerates downward at a, it creates a counterclockwise angular acceleration A of the drum. The moment of inertia of the drum is

(a)

IO = mk2O = (60 kg)(0.25 m)2 = 3.75 kg # m2

There are five unknowns, namely Ox , Oy , T, a, and a. 60 (9.81) N

0.4 m

O Ox

T

17

equations of Motion. Applying the translational equations of motion Fx = m(aG)x and Fy = m(aG)y to the drum is of no consequence to the solution, since these equations involve the unknowns Ox and Oy . Thus, for the drum and block, respectively, a + MO = IOa;

y Oy

T

A

x

+ c Fy = m(aG)y;

T(0.4 m) = (3.75 kg # m2)a

(1)

-20(9.81)N + T = -(20 kg)a

(2)

Kinematics. Since the point of contact A between the cord and drum has a tangential component of acceleration a, Fig. 17–16a, then

a

a +a = ar; (b)

T = 106 N a = 4.52 m>s2 a = 11.3 rad>s2d

60 (9.81) N O Ox

(3)

Solving the above equations,

20 (9.81) N

0.4 m

a = a(0.4 m)

=

IOa 0.4 m O

Oy

20(9.81) N

(20 kg)a (c)

Fig. 17–16

Ans. Solution ii Free-body and Kinetic Diagrams. The cable tension T can be eliminated from the analysis by considering the drum and block as a single system, Fig. 17–16c. The kinetic diagram is shown since moments will be summed about point O. equations of Motion. Using Eq. 3 and applying the moment equation about O to eliminate the unknowns Ox and Oy , we have a + M O = (k )O; [20(9.81) N] (0.4 m) = # (3.75 kg m2)a + [20 kg(a 0.4 m)](0.4 m) a = 11.3 rad>s2 Ans. NoTE: If the block were removed and a force of 20(9.81) N were applied to the cord, show that a = 20.9 rad>s2. This value is larger since the block has an inertia, or resistance to acceleration.

17.4

ExamplE

equations of Motion: rotation aBout a fixed axis

447

17.11

The slender rod shown in Fig. 17–17a has a mass m and length l and is released from rest when u = 0. Determine the horizontal and vertical components of force which the pin at A exerts on the rod at the instant u = 90.

A u

Solution Free-body and Kinetic Diagrams. The free-body diagram for the rod in the general position u is shown in Fig. 17–17b. For convenience, the force components at A are shown acting in the n and t directions. Note that A acts clockwise and so (aG)t acts in the +t direction. The moment of inertia of the rod about point A is IA = 13 ml2. equations of Motion. to eliminate An and At. 2

(a)

Moments will be summed about A in order 2

+ aFn = mv rG; + bFt = marG; c + MA = IAa;

l

An - mg sin u = mv (l>2) At + mg cos u = ma(l>2) mg cos u(l>2) = 1 13 ml2 2 a

An

(1) (2) (3)

A u

At

–l 2

Kinematics. For a given angle u there are four unknowns in the above three equations: An , At , v, and a. As shown by Eq. 3, a is not constant; rather, it depends on the position u of the rod. The necessary fourth equation is obtained using kinematics, where a and v can be related to u by the equation

u

(4)

v dv = a du

Note that the positive clockwise direction for this equation agrees with that of Eq. 3. This is important since we are seeking a simultaneous solution. In order to solve for v at u = 90, eliminate a from Eqs. 3 and 4, which yields

17 mg

=

(c +)

G

mv2

( (

mÅ –l 2

G

( –2l ( IGÅ

v dv = (1.5g>l) cos u du Since v = 0 at u = 0, we have L0

(b)

90

v

v dv = (1.5g>l)

L0

cos u du

v2 = 3g>l Substituting this value into Eq. 1 with u = 90 and solving Eqs. 1 to 3 yields a = 0 At = 0 A n = 2.5 mg Ans. NoTE: If M A = (k )A is used, one must account for the moments

of IG A and m(aG)t about A.

Fig. 17–17

448

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

FunDaMentaL prObLeMs F17–7. The [email protected] wheel has a radius of gyration about its center O of kO = 500 mm. If the wheel starts from rest, determine its angular velocity in t = 3 s.

0.6 m

P  100 N

F17–10. At the instant shown, the [email protected] disk has a counterclockwise angular velocity of v = 10 rad>s. Determine the tangential and normal components of reaction of the pin O on the disk and the angular acceleration of the disk at this instant. O

v  10 rad/s 5 4

O

3

P  50 N

0.3 m

Prob. F17–7

17

F17–8. The [email protected] disk is subjected to the couple moment of M = (9t) N # m, where t is in seconds. Determine the angular velocity of the disk when t = 4 s starting from rest.

Prob. F17–10 F17–11. The uniform slender rod has a mass of 15 kg. Determine the horizontal and vertical components of reaction at the pin O, and the angular acceleration of the rod just after the cord is cut.

O 0.3 m O

M  (9t) Nm 0.6 m

Prob. F17–8 F17–9. At the instant shown, the uniform [email protected] slender rod has a counterclockwise angular velocity of v = 6 rad>s. Determine the tangential and normal components of reaction of pin O on the rod and the angular acceleration of the rod at this instant.

0.3 m

Prob. F17–11 F17–12. The uniform [email protected] slender rod is being pulled by the cord that passes over the small smooth peg at A. If the rod has a counterclockwise angular velocity of v = 6 rad>s at the instant shown, determine the tangential and normal components of reaction at the pin O and the angular acceleration of the rod. P  300 N A

0.8 m 0.3 m

0.6 m O

M  60 Nm

v  6 rad/s O 0.6 m

Prob. F17–9

Prob. F17–12

0.3 m

17.4

449

equations of Motion: rotation aBout a fixed axis

prObLeMs 17–57. The 10-kg wheel has a radius of gyration k A = 200 mm. If the wheel is subjected to a moment M = (5t) N # m, where t is in seconds, determine its angular velocity when t = 3 s starting from rest. Also, compute the reactions which the fixed pin A exerts on the wheel during the motion.

*17–60. The bent rod has a mass of 2 kg>m. If it is released from rest in the position shown, determine its initial angular acceleration and the horizontal and vertical components of reaction at A.

M

1.5 m A

C

Prob. 17–57

A

1.5 m

17–58. The uniform 24-kg plate is released from rest at the position shown. Determine its initial angular acceleration and the horizontal and vertical reactions at the pin A. B

A

17

Prob. 17–60

0.5 m

17–61. If a horizontal force of P = 100 N is applied to the 300-kg reel of cable, determine its initial angular acceleration. The reel rests on rollers at A and B and has a radius of gyration of k O = 0.6 m.

0.5 m

Prob. 17–58 17–59. The uniform slender rod has a mass m. If it is released from rest when u = 0, determine the magnitude of the reactive force exerted on it by pin B when u = 90.

A

P

L 3

0.75 m O

B

1m u

2 L 3

20 A

20 B

C

Prob. 17–59

Prob. 17–61

450

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–62. The 20-kg roll of paper has a radius of gyration kA = 90 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. If the roll rests against a wall for which the coefficient of kinetic friction is μk = 0.2 and a vertical force F = 30 N is applied to the end of the paper, determine the angular acceleration of the roll as the paper unrolls.

and

a C C e l e r at i o n

*17–64. The kinetic diagram representing the general rotational motion of a rigid body about a fixed axis passing through O is shown in the figure. Show that IGA may be eliminated by moving the vectors m(aG)t and m(aG)n to 2 point P, located a distance rGP = k G >rOG from the center of mass G of the body. Here k G represents the radius of gyration of the body about an axis passing through G. The point P is called the center of percussion of the body.

B

300 mm

a P A

C

m(aG)t

125 mm

G m(aG)n

IG a rGP

O rOG

F

17

Prob. 17–62 17–63. The 20-kg roll of paper has a radius of gyration kA = 90 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. If the roll rests against a wall for which the coefficient of kinetic friction is μk = 0.2, determine the constant vertical force F that must be applied to the roll to pull off 1 m of paper in t = 3 s starting from rest. Neglect the mass of paper that is removed. B

Prob. 17–64

17–65. Gears A and B have a mass of 50 kg and 15 kg, respectively. Their radii of gyration about their respective centers of mass are kC = 250 mm and kD = 150 mm. If a torque of M = 200(1 - e-0.2t) N # m, where t is in seconds, is applied to gear A, determine the angular velocity of both gears when t = 3 s, starting from rest.

300 mm

C

M

200(1

300 mm

A 125 mm

e

0.2t

200 mm D

C

B F

Prob. 17–63

)N m

A

Prob. 17–65

17.4

451

equations of Motion: rotation aBout a fixed axis

17–66. The reel of cable has a mass of 400 kg and a radius of gyration of kA = 0.75 m. Determine its angular velocity when t = 2 s, starting from rest, if the force P = (20t2 + 80) N, when t is in seconds. Neglect the mass of the unwound cable, and assume it is always at a radius of 0.5 m.

17–69. If the cord at B suddenly fails, determine the horizontal and vertical components of the initial reaction at the pin A, and the angular acceleration of the 120-kg beam. Treat the beam as a uniform slender rod. 800 N

0.5 m

P

B

1m

A A

2m

2m

Prob. 17–69

Prob. 17–66 17–67. The door will close automatically using torsional springs mounted on the hinges. Each spring has a stiffness k = 50 N # m>rad so that the torque on each hinge is M = (50u) N # m, where u is measured in radians. If the door is released from rest when it is open at u = 90, determine its angular velocity at the instant u = 0. For the calculation, treat the door as a thin plate having a mass of 70 kg.

17–70. The device acts as a pop-up barrier to prevent the passage of a vehicle. It consists of a 100-kg steel plate AC and a 200-kg counterweight solid concrete block located as shown. Determine the moment of inertia of the plate and block about the hinged axis through A. Neglect the mass of the supporting arms AB. Also, determine the initial angular acceleration of the assembly when it is released from rest at u = 45°. 17 0.3 m

*17–68. The door will close automatically using torsional springs mounted on the hinges. If the torque on each hinge is M = ku, where u is measured in radians, determine the required torsional stiffness k so that the door will close (u = 0) with an angular velocity v = 2 rad>s when it is released from rest at u = 90. For the calculation, treat the door as a thin plate having a mass of 70 kg.

θ

B

0.4 m M 1.5 m

0.5 m 0.5 m C

A 1.25 m

Prob. 17–70 17–71. A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = ( ¼ u2 ) N, where u is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revolutions. The disk has an initial angular velocity of v0 = 1 rad>s.

F A

0.4 m

300 mm

M

O 1.2 m

Probs. 17–67/68

B u

Prob. 17–71

v

452

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

*17–72. Block A has a mass m and rests on a surface having a coefficient of kinetic friction μk. The cord attached to A passes over a pulley at C and is attached to a block B having a mass 2m. If B is released, determine the acceleration of A. Assume that the cord does not slip over the pulley. The pulley can be approximated as a thin disk of radius r and 1 mass m. Neglect the mass of the cord. 4

r

A

and

a C C e l e r at i o n

17–75. The 30-kg disk is originally spinning at v = 125 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is μC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of force which the member AB exerts on the pin at A during this time? Neglect the mass of AB.

C

0.5 m

B 0.3 m

B

Prob. 17–72

0.5 m

A

17 17–73. The two blocks A and B have a mass of 5 kg and 10 kg, respectively. If the pulley can be treated as a disk of mass 3 kg and radius 0.15 m, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley. 17–74. The two blocks A and B have a mass mA and mB, respectively, where mB 7 mA. If the pulley can be treated as a disk of mass M, determine the acceleration of block A. Neglect the mass of the cord and any slipping on the pulley.

v  125 rad/s

C

Prob. 17–75

*17–76. The wheel has a mass of 25 kg and a radius of gyration kB = 0.15 m. It is originally spinning at v = 40 rad>s. If it is placed on the ground, for which the coefficient of kinetic friction is μC = 0.5, determine the time required for the motion to stop. What are the horizontal and vertical components of reaction which the pin at A exerts on AB during this time? Neglect the mass of AB.

0.4 m r O

A 0.3 m B

0.2 m

A

V

B C

Probs. 17–73/74

Prob. 17–76

17.4

17–77. The disk has a mass of 20 kg and is originally spinning at the end of the strut with an angular velocity of v = 60 rad>s. If it is then placed against the wall, for which the coefficient of kinetic friction is μk = 0.3 determine the time required for the motion to stop. What is the force in strut BC during this time?

B

453

equations of Motion: rotation aBout a fixed axis

17–79. Cable is unwound from a spool supported on small rollers at A and B by exerting a force T = 300 N on the cable. Compute the time needed to unravel 5 m of cable from the spool if the spool and cable have a total mass of 600 kg and a radius of gyration of kO = 1.2 m. For the calculation, neglect the mass of the cable being unwound and the mass of the rollers at A and B. The rollers turn with no friction.

A

150 mm T  300 N

1.5 m 30

0.8 m O

A

60

B

C 1m

17

Prob. 17–79

Prob. 17–77

17–78. The 5-kg cylinder is initially at rest when it is placed in contact with the wall B and the rotor at A. If the rotor always maintains a constant clockwise angular velocity v = 6 rad>s, determine the initial angular acceleration of the cylinder. The coefficient of kinetic friction at the contacting surfaces B and C is μk = 0.2.

*17–80. The 20-kg roll of paper has a radius of gyration kA = 120 mm about an axis passing through point A. It is pin supported at both ends by two brackets AB. The roll rests on the floor, for which the coefficient of kinetic friction is μk = 0.2. If a horizontal force F = 60 N is applied to the end of the paper, determine the initial angular acceleration of the roll as the paper unrolls.

F B 125 mm

A v 45

A

300 mm B

C

C 400 mm

Prob. 17–78

Prob. 17–80

454

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–81. The armature (slender rod) AB has a mass of 0.2 kg and can pivot about the pin at A. Movement is controlled by the electromagnet E, which exerts a horizontal attractive force on the armature at B of FB = (0.2(10-3)l-2) N, where l in meters is the gap between the armature and the magnet at any instant. If the armature lies in the horizontal plane, and is originally at rest, determine the speed of the contact at B the instant l = 0.01 m. Originally l = 0.02 m.

and

a C C e l e r at i o n

17–83. The bar has a weight per length of w. If it is rotating in the vertical plane at a constant rate v about point O, determine the internal normal force, shear force, and moment as a function of x and u.

O v l

B

u L

150 mm

E

x

Prob. 17–83 17

A

*17–84. Determine the angular acceleration of the 25-kg diving board and the horizontal and vertical components of reaction at the pin A the instant the man jumps off. Assume that the board is uniform and rigid, and that at the instant he jumps off the spring is compressed a maximum amount of 200 mm, v = 0, and the board is horizontal. Take k = 7 kN>m.

Prob. 17–81

17–82. The 4-kg slender rod is initially supported horizontally by a spring at B and pin at A. Determine the angular acceleration of the rod and the acceleration of the rod’s mass center at the instant the 100-N force is applied. 1.5 m

1.5 m

A k 100 N 1.5 m

1.5 m

A

B k  20 N/m

Prob. 17–82

Prob. 17–84

17.4

455

equations of Motion: rotation aBout a fixed axis

17–85. The lightweight turbine consists of a rotor which is powered from a torque applied at its center. At the instant the rotor is horizontal it has an angular velocity of 15 rad>s and a clockwise angular acceleration of 8 rad>s2. Determine the internal normal force, shear force, and moment at a section through A. Assume the rotor is a 50-m-long slender rod, having a mass of 3 kg>m.

10 m

17–87. The 100-kg pendulum has a center of mass at G and a radius of gyration about G of kG = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant u = 90° when the pendulum is rotating at v = 8 rad>s. Neglect the weight of the beam and the support. *17–88. The 100-kg pendulum has a center of mass at G and a radius of gyration about G of kG = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant u = 0° when the pendulum is rotating at v = 4 rad>s. Neglect the weight of the beam and the support.

A C

25 m

u

0.75 m v

1m G

A 0.6 m

Prob. 17–85

l

0.6 m

Probs. 17–87/88

17–86. The two-bar assembly is released from rest in the position shown. Determine the initial bending moment at the fixed joint B. Each bar has a mass m and length l.

A

B

17–89. The “Catherine wheel” is a firework that consists of a coiled tube of powder which is pinned at its center. If the powder burns at a constant rate of 20 g>s such that the exhaust gases always exert a force having a constant magnitude of 0.3 N, directed tangent to the wheel, determine the angular velocity of the wheel when 75% of the mass is burned off. Initially, the wheel is at rest and has a mass of 100 g and a radius of r = 75 mm. For the calculation, consider the wheel to always be a thin disk.

B

r C 0.3 N

l

C

Prob. 17–86

Prob. 17–89

17

456

Chapter 17

planar KinetiCs

F3

17.5 V aG

M1 F4

G

A

M2 F2

F1

of a

(a)

and

a C C e l e r at i o n

Equations of Motion: General Plane Motion

The rigid body (or slab) shown in Fig. 17–18a is subjected to general plane motion caused by the externally applied force and couple-moment system. The free-body and kinetic diagrams for the body are shown in Fig. 17–18b. If an x and y inertial coordinate system is established as shown, the three equations of motion are

Fx = m(aG)x Fy = m(aG)y MG = IGa

y

F3

rigid Body: forCe

(17–17)

x M1 F4

G M2

W F2

In some problems it may be convenient to sum moments about a point P other than G in order to eliminate as many unknown forces as possible from the moment summation. When used in this more general case, the three equations of motion are

F1



17

Fx = m(aG)x Fy = m(aG)y MP = (mk)P maG

m(aG)y G

(17–18)

m(aG)x

Here (mk)P represents the moment sum of IG A and maG (or its components) about P as determined by the data on the kinetic diagram.

IGA

Moment Equation About the IC. There is a particular type of (b)

Fig. 17–18 A F

problem that involves a uniform disk, or body of circular shape, that rolls on a rough surface without slipping, Fig. 17–19. If we sum the moments about the instantaneous center of zero velocity, then (mk)IC becomes IICa, so that

MIC = IICa

IC

Fig. 17–19

(17–19)

This result compares withMO = IOa, which is used for a body pinned at point O, Eq. 17–16. See Prob. 17–90.

17.5

457

equations of Motion: general plane Motion

W Gx

G Gy

A

FA

=

NA

IG A

procedure for analysis Kinetic problems involving general plane motion of a rigid body can be solved using the following procedure. Free-Body Diagram. • Establish the x, y inertial coordinate system and draw the freebody diagram for the body. • Specify the direction and sense of the acceleration of the mass center, aG , and the angular acceleration A of the body. • Determine the moment of inertia IG . • Identify the unknowns in the problem. • If it is decided that the rotational equation of motion MP = (mk)P is to be used, then consider drawing the kinetic diagram in order to help “visualize” the “moments” developed by the components m(aG)x , m(aG)y , and IG A when writing the terms in the moment sum (mk)P . Equations of motion. • Apply the three equations of motion in accordance with the established sign convention. • When friction is present, there is the possibility for motion with no slipping or tipping. Each possibility for motion should be considered. Kinematics. • Use kinematics if a complete solution cannot be obtained strictly from the equations of motion. • If the body’s motion is constrained due to its supports, additional equations may be obtained by using aB = aA + aB>A , which relates the accelerations of any two points A and B on the body.

• When a wheel, disk, cylinder, or ball rolls without slipping, then aG = ar.

G maG

d A

As the soil compactor, or “sheep’s foot roller” moves forward, the roller has general plane motion. The forces shown on its free-body diagram cause the effects shown on the kinetic 17 diagram. If moments are summed about the mass center, G, then MG = IGa. However, if moments are summed about point A (the IC) then a + MA = IGa + (maG)d = IAa.

458

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

17.12

ExamplE

Determine the angular acceleration of the spool in Fig. 17–20a. The spool has a mass of 8 kg and a radius of gyration of kG = 0.35 m. The cords of negligible mass are wrapped around its inner hub and outer rim.

100 N

G

A 0.5 m

0.2 m

(a)

equations of Motion. + c Fy = m(aG)y; T + 100 N - 78.48 N = (8 kg)aG c + MG = IGa;

100 N

G 0.5 m

A

100 N(0.2 m) - T(0.5 m) = (0.980 kg # m )a

(c +) aG = ar;

a = 10.3 rad>s2 aG = 5.16 m>s2 T = 19.8 N

=

c + MA = (mk)A;

0.5 m

A

100 N(0.7 m) - 78.48 N(0.5 m)

= (0.980 kg # m2)a + [(8 kg)aG](0.5 m)

Using Eq. (3), a = 10.3 rad>s2

Ans.

Solution iii equations of Motion. The simplest way to solve this problem is to realize that point A is the IC for the spool. Then Eq. 17–19 applies. c + MA = IAa;

Fig. 17–20

Ans.

Solution ii equations of Motion. We can eliminate the unknown T by summing moments about point A. From the free-body and kinetic diagrams Figs. 17–20b and 17–20c, we have

(8 kg) aG

(b)

(3)

aG = a (0.5 m)

78.48 N

G

(2)

Solving Eqs. 1 to 3, we have

0.2 m

(0.980 kgm2) A

(1)

2

Kinematics. A complete solution is obtained if kinematics is used to relate aG to a. In this case the spool “rolls without slipping” on the cord at A. Hence, we can use the results of Example 16.4 or 16.15 so that,

T

17

Solution i Free-body and Kinetic Diagrams. Fig. 17–20b. The 100-N force causes aG to act upward. Also, A acts clockwise, since the spool winds around the cord at A. There are three unknowns T, aG , and a. The moment of inertia of the spool about its mass center is IG = mk2G = 8 kg(0.35 m)2 = 0.980 kg # m2

(100 N)(0.7 m) - (78.48 N)(0.5 m) = [0.980 kg # m2 + (8 kg)(0.5 m)2]a a = 10.3 rad>s2

17.5

ExamplE

17.13

The 25-kg wheel shown in Fig. 17–21a has a radius of gyration kG = 0.2 m. If a 50 N # m couple moment is applied to the wheel, determine the acceleration of its mass center G. The coefficients of static and kinetic friction between the wheel and the plane at A are ms = 0.3 and mk = 0.25, respectively. Solution Free-Body Diagram. By inspection of Fig. 17–21b, it is seen that the couple moment causes the wheel to have a clockwise angular acceleration of A. As a result, the acceleration of the mass center, aG , is directed to the right. The moment of inertia is IG = mk2G = 25 kg (0.2 m)2 = 1.0 kg # m2

A (a)

25(9.81)N

y

A

aG x

0.36 m

+ F = m(a ) ; S x G x

FA = (25 kg) aG

(1)

+ c Fy = m(aG)y;

NA - 25(9.81)N = 0

(2)

50 N # m - 0.36 m(FA) = (1.0 kg # m2)a

(3)

Kinematics (No Slipping).

If this assumption is made, then

aG = (0.36 m)a

(4)

Solving Eqs. 1 to 4, NA = 245.25 N a = 11.79 rad>s

FA = 106.1 N 2

aG = 4.245 m>s2

This solution requires that no slipping occurs, i.e., FA … msNA . However, since 106.1 N 7 0.3(25)(9.81)N = 73.6 N, the wheel slips as it rolls. Equation 4 is not valid, and so FA = mkNA , or FA = 0.25NA

(5)

Solving Eqs. 1 to 3 and 5 yields NA = 245.25 N FA = 61.31 N a = 27.93 rad>s2 aG = 2.45 m>s2 S

FA NA

17 (b)

Fig. 17–21

A fourth equation is needed for a complete solution.

(c +)

G 0.36 m

G

Equations of Motion.

c + MG = IGa;

M  50 Nm

50 N m

The unknowns are NA , FA , aG , and a.

(Slipping).

459

Equations of Motion: GEnEral PlanE Motion

Ans.

460

Chapter 17

planar KinetiCs

rigid Body: forCe

and

a C C e l e r at i o n

17.14

ExamplE

The uniform slender pole shown in Fig. 17–22a has a mass of 100 kg. If the coefficients of static and kinetic friction between the end of the pole and the surface are ms = 0.3, and mk = 0.25, respectively, determine the pole’s angular acceleration at the instant the 400-N horizontal force is applied. The pole is originally at rest.

3m

400 N 0.5 m

A (a)

1.5 m

17

of a

1m

Solution Free-body and Kinetic Diagrams. Figure 17–22b. The path of motion of the mass center G will be along an unknown curved path having a radius of curvature r, which is initially on a vertical line. However, there is no normal or y component of acceleration since the pole is originally at rest, i.e., vG = 0, so that (aG)y = v2G >r = 0. We will assume the mass center accelerates to the right and that the pole has a clockwise angular acceleration of A. The unknowns are NA , FA , aG , and a.

G

equation of Motion.

981 N

+ F = m(a ) ; S x G x + c Fy = m(aG)y;

400 N

c + MG = IGa; FA(1.5 m) - (400 N)(1 m) =

FA

(1)

400 N - FA = (100 kg)aG NA - 981 N = 0 1 [12 (100

(2) kg)(3 m)2]a

(3)

A fourth equation is needed for a complete solution.

NA

=

Kinematics (no slipping). With this assumption, point A acts as a “pivot” so that a is clockwise, then aG is directed to the right. aG = (1.5 m) a

aG = arAG;

(4)

Solving Eqs. 1 to 4 yields (100 kg)aG

NA = 981 N aG = 1 m>s

IGA

2

FA = 300 N a = 0.667 rad>s2

The assumption of no slipping requires FA … msNA . However, 300 N 7 0.3(981 N) = 294 N and so the pole slips at A. (b)

Fig. 17–22

(slipping). For this case Eq. 4 does not apply. Instead the frictional equation FA = mkNA must be used. Hence, FA = 0.25NA

(5)

Solving Eqs. 1 to 3 and 5 simultaneously yields NA = 981 N FA = 245 N aG = 1.55 m>s2 a = -0.428 rad>s2 = 0.428 rad>s2d

Ans.

17.5

ExamplE

461

equations of Motion: general plane Motion

17.15

The uniform 50-kg bar in Fig. 17–23a is held in the equilibrium position by cords AC and BD. Determine the tension in BD and the angular acceleration of the bar immediately after AC is cut. Solution Free-body and Kinetic Diagrams. unknowns, TB , (aG)x , (aG)y , and a.

Fig. 17–23b. There are four

C

D

A

B 3m (a)

equations of Motion. + Fx = m(aG)x; S

0 = 50 kg (aG)x

50(9.81) N

TB

(aG)x = 0 + c Fy = m(aG)y; a + MG = IGa;

TB - 50(9.81)N = -50 kg (aG)y TB(1.5 m) = J

1 (50 kg)(3 m)2 R a 12

G

(1)

B 1.5 m

(2)

17

Kinematics. Since the bar is at rest just after the cable is cut, then its angular velocity and the velocity of point B at this instant are equal to zero. Thus (aB)n = v2B >rBD = 0. Therefore, aB only has a tangential component, which is directed along the x axis, Fig. 17–23c. Applying the relative acceleration equation to points G and B,

IGA (50 kg)(aG)x (50 kg)(aG)y (b)

aG = aB + A * rG/B - v2rG/B - (aG)yj = aBi + (ak) * ( -1.5i) - 0 -(aG)yj = aBi - 1.5aj

(aG)y G (aG)x  0

Equating the i and j components of both sides of this equation, 0 

0 = aB (3)

(aG)y = 1.5a

1.5 m (c)

Fig. 17–23

Solving Eqs. (1) through (3) yields a = 4.905 rad>s2

Ans. Ans.

TB = 123 N (aG)y = 7.36 m>s

rG/B

2

B

aB

462

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

FunDaMentaL prObLeMs F17–13. The uniform [email protected] slender bar is initially at rest on a smooth horizontal plane when the forces are applied. Determine the acceleration of the bar’s mass center and the angular acceleration of the bar at this instant. 20 N 0.75 m

F17–16. The [email protected] sphere rolls down the inclined plane without slipping. Determine the angular acceleration of the sphere and the acceleration of its mass center.

0.5 m

1.75 m

0.15 m

30

80 N

Prob. F17–13

Prob. F17–16

F17–14. The [email protected] cylinder rolls without slipping on the horizontal plane. Determine the acceleration of its mass center and its angular acceleration.

F17–17. The [email protected] spool has a radius of gyration about its mass center of kG = 300 mm. If the couple moment is applied to the spool and the coefficient of kinetic friction between the spool and the ground is mk = 0.2, determine the angular acceleration of the spool, the acceleration of G and the tension in the cable.

17 0.3 m P  200 N

0.4 m B

A 0.6 m G

M  450 Nm

Prob. F17–14 F17–15. The [email protected] wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it slips as it rolls. Determine the angular acceleration of the wheel and the acceleration of the wheel’s center O. The coefficient of kinetic friction between the wheel and the plane is mk = 0.5. 0.4 m M  100 Nm

Prob. F17–17 F17–18. The [email protected] slender rod is pinned to a small roller A that slides freely along the slot. If the rod is released from rest at u = 0, determine the angular acceleration of the rod and the acceleration of the roller immediately after the release. A u

O 0.6 m

Prob. F17–15

Prob. F17–18

17.5

equations of Motion: general plane Motion

463

prObLeMs 17–90. If the disk in Fig. 17–19 rolls without slipping, show that when moments are summed about the instantaneous center of zero velocity, IC, it is possible to use the moment equation M IC = IIC a, where IIC represents the moment of inertia of the disk calculated about the instantaneous axis of zero velocity. 17–91. The slender 12-kg bar has a clockwise angular velocity of v = 2 rad>s when it is in the position shown. Determine its angular acceleration and the normal reactions of the smooth surface A and B at this instant.

17–93. The spool has a mass of 500 kg and a radius of gyration kG = 1.30 m. It rests on the surface of a conveyor belt for which the coefficient of static friction is ms = 0.5 and the coefficient of kinetic friction is mk = 0.4. If the conveyor accelerates at aC = 1 m>s2 , determine the initial tension in the wire and the angular acceleration of the spool. The spool is originally at rest. 17–94. The spool has a mass of 500 kg and a radius of gyration kG = 1.30 m. It rests on the surface of a conveyor belt for which the coefficient of static friction is ms = 0.5. Determine the greatest acceleration aC of the conveyor so that the spool will not slip. Also, what are the initial tension in the wire and the angular acceleration of the spool? The spool is originally at rest.

B 0.8 m

1.6 m G

aC

3m

17

Probs. 17–93/94 60

17–95. The 20-kg punching bag has a radius of gyration about its center of mass G of k G = 0.4 m. If it is initially at rest and is subjected to a horizontal force F = 30 N, determine the initial angular acceleration of the bag and the tension in the supporting cable AB.

A

Prob. 17–91

*17–92. The 2-kg slender bar is supported by cord BC and then released from rest at A. Determine the initial angular acceleration of the bar and the tension in the cord.

A 1m B

C

0.3 m

30° B

A

G

0.6 m F

300 mm

Prob. 17–92

Prob. 17–95

464

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

*17–96. The assembly consists of an 8-kg disk and a 10-kg bar which is pin connected to the disk. If the system is released from rest, determine the angular acceleration of the disk. The coefficients of static and kinetic friction between the disk and the inclined plane are ms = 0.6 and mk = 0.4, respectively. Neglect friction at B.

and

a C C e l e r at i o n

*17–100. Wheel C has a mass of 60 kg and a radius of gyration of 0.4 m, whereas wheel D has a mass of 40 kg and a radius of gyration of 0.35 m. Determine the angular acceleration of each wheel at the instant shown. Neglect the mass of the link and assume that the assembly does not slip on the plane.

17–97. Solve Prob. 17–96 if the bar is removed. The coefficients of static and kinetic friction between the disk and inclined plane are ms = 0.15 and mk = 0.1, respectively.

0.1 m

B 2m

D 0.5 m

B

A

1m 0.1 m

A 0.3 m

C 0.5 m

C 30

30

Probs. 17–96/97

Prob. 17–100

17–98. A force of F = 10 N is applied to the 10-kg ring as shown. If slipping does not occur, determine the ring’s initial angular acceleration, and the acceleration of its mass center, G. Neglect the thickness of the ring.

17–101. The spool has a mass of 100 kg and a radius of gyration of k G = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 50 N.

17–99. If the coefficient of static friction at C is μs = 0.3, determine the largest force F that can be applied to the 5-kg ring, without causing it to slip. Neglect the thickness of the ring.

17–102. Solve Prob. 17–101 if the cord and force P = 50 N are directed vertically upwards.

17

F

A G

30

17–103. The spool has a mass of 100 kg and a radius of gyration k G = 0.3 m. If the coefficients of static and kinetic friction at A are ms = 0.2 and mk = 0.15, respectively, determine the angular acceleration of the spool if P = 600 N.

45 P 250 mm

G

400 mm

0.4 m

C

Probs. 17–98/99

A

Probs. 17–101/102/103

17.5 *17–104. The uniform bar of mass m and length L is balanced in the vertical position when the horizontal force P is applied to the roller at A. Determine the bar’s initial angular acceleration and the acceleration of its top point B. 17–105. Solve Prob. 17–104 if the roller is removed and the coefficient of kinetic friction at the ground is μk.

equations of Motion: general plane Motion

465

17–107. The 500-kg concrete culvert has a mean radius of 0.5 m. If the truck has an acceleration of 3 m>s2, determine the culvert’s angular acceleration. Assume that the culvert does not slip on the truck bed, and neglect its thickness.

3 m/s2

4m 0.5m

B

Prob. 17–107 *17–108. The 12-kg uniform bar is supported by a roller at A. If a horizontal force of F = 80 N is applied to the roller, determine the acceleration of the center of the roller at the instant the force is applied. Neglect the weight and the size of the roller.

L

P

A A

F  80 N

17

Probs. 17–104/105

17–106. A “lifted” truck can become a road hazard since the bumper is high enough to ride up a standard car in the event the car is rear-ended. As a model of this case consider the truck to have a mass of 2.70 Mg, a mass center G, and a radius of gyration about G of kG = 1.45 m. Determine the horizontal and vertical components of acceleration of the mass center G, and the angular acceleration of the truck, at the moment its front wheels at C have just left the ground and its smooth front bumper begins to ride up the back of the stopped car so that point B has a velocity of vB = 8 m>s at 20º from the horizontal. Assume the wheels are free to roll, and neglect the size of the wheels and the deformation of the material.

20

G

B

1.3 m

2m

Prob. 17–108 17–109. The semicircular disk having a mass of 10 kg is rotating at v = 4 rad>s at the instant u = 60. If the coefficient of static friction at A is ms = 0.5, determine if the disk slips at this instant.

v 0.4 m

O

G A 1.6 m

C 0.4 m 1.2 m

Prob. 17–106

u A

Prob. 17–109

4 (0.4) ——— m 3p

466

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

17–110. The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the initial angular acceleration of the disk and the acceleration of its mass center. The coefficient of kinetic friction between the disk and the floor is μk.

and

a C C e l e r at i o n

17–113. The 30-kg uniform slender rod AB rests in the position shown when the couple moment of M = 150 N # m is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers.

17–111. The uniform disk of mass m is rotating with an angular velocity of v0 when it is placed on the floor. Determine the time before it starts to roll without slipping. What is the angular velocity of the disk at this instant? The coefficient of kinetic friction between the disk and the floor is μk.

A

0.75 m

M

150 N m

0.75 m

v0

B

r

Prob. 17–113 17–114. The 30-kg slender rod AB rests in the position shown when the horizontal force P = 50 N is applied. Determine the initial angular acceleration of the rod. Neglect the mass of the rollers..

Probs. 17–110/111 17

A

*17–112. The 20-kg disk A is attached to the 10-kg block B using the cable and pulley system shown. If the disk rolls without slipping, determine its angular acceleration and the acceleration of the block when they are released. Also, what is the tension in the cable? Neglect the mass of the pulleys.

1.5 m

B P

50 N

Prob. 17–114

A

G 0.2 m

17–115. The solid ball of radius r and mass m rolls without slipping down the 60° trough. Determine its angular acceleration.

30 30

B

Prob. 17–112

45

Prob. 17–115

17.5 *17–116. A cord is wrapped around each of the two 10-kg disks. If they are released from rest determine the angular acceleration of each disk and the tension in the cord C. Neglect the mass of the cord.

467

Equations of Motion: GEnEral PlanE Motion

17–118. A long strip of paper is wrapped into two rolls, each having a mass of 8 kg. Roll A is pin supported about its center whereas roll B is not centrally supported. If B is brought into contact with A and released from rest, determine the initial tension in the paper between the rolls and the angular acceleration of each roll. For the calculation, assume the rolls to be approximated by cylinders.

A

D 90 mm 90 mm

B

A 90 mm

Prob. 17–118 17–119. The uniform beam has a weight W. If it is originally at rest while being supported at A and B by cables, determine the tension in cable A if cable B suddenly fails. Assume the beam is a slender rod.

C

B

17

90 mm

Prob. 17–116

A

17–117. The disk of mass m and radius r rolls without slipping on the circular path. Determine the normal force which the path exerts on the disk and the disk’s angular acceleration if at the instant shown the disk has an angular velocity of V.

B

L –– 4

L –– 2

L –– 4

Prob. 17–119 *17–120. By pressing down with the finger at B, a thin ring having a mass m is given an initial velocity v0 and a backspin V0 when the finger is released. If the coefficient of kinetic friction between the table and the ring is μk, determine the distance the ring travels forward before backspinning stops.

B v0

u R v r

v0 r

A

Prob. 17–117

Prob. 17–120

468

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

COnCeptuaL prObLeMs C17–1. The truck is used to pull the heavy container. To be most effective at providing traction to the rear wheels at A, is it best to keep the container where it is or place it at the front of the trailer? Use appropriate numerical values to explain your answer.

C17–3. How can you tell the driver is accelerating this SUV? To explain your answer, draw the free-body and kinetic diagrams. Here power is supplied to the rear wheels. Would the photo look the same if power were supplied to the front wheels? Will the accelerations be the same? Use appropriate numerical values to explain your answers.

A

17 Prob. C17–1

Prob. C17–3

C17–2. The tractor is about to tow the plane to the right. Is it possible for the driver to cause the front wheel of the plane to lift off the ground as he accelerates the tractor? Draw the free-body and kinetic diagrams and explain algebraically (letters) if and how this might be possible.

C17–4. Here is something you should not try at home, at least not without wearing a helmet! Draw the free-body and kinetic diagrams and show what the rider must do to maintain this position. Use appropriate numerical values to explain your answer.

Prob. C17–2

Prob. C17–4

469

Chapter review

Chapter review Moment of Inertia The moment of inertia is a measure of the resistance of a body to a change in its angular velocity. It is defined by I = 1 r2dm and will be different for each axis about which it is computed.

m G r

Many bodies are composed of simple shapes. If this is the case, then tabular values of I can be used, such as the ones given on the inside back cover of this book. To obtain the moment of inertia of a composite body about any specified axis, the moment of inertia of each part is determined about the axis and the results are added together. Doing this often requires use of the parallel-axis theorem.

dm d

I = IG + md2

I

IG

17

Planar Equations of Motion

Fx = m(aG)x

Fn = m(aG)n

The equations of motion define the translational, and rotational motion of a rigid body. In order to account for all of the terms in these equations, a free-body diagram should always accompany their application, and for some problems, it may also be convenient to draw the kinetic diagram which shows maG and IGA.

Fy = m(aG)y

Ft = m(aG)t

MG = 0 Rectilinear translation

Fn = m(aG)n = mv2rG Ft = m(aG)t = marG MG = IGa or MO = IOa Rotation About a Fixed Axis Fx = m(aG)x Fy = m(aG)y MG = IGa or MP = (mk)P General Plane Motion

MG = 0 Curvilinear translation

470

Chapter 17

planar KinetiCs

of a

rigid Body: forCe

and

a C C e l e r at i o n

review prObLeMs R17–1. The handcart has a mass of 200 kg and center of mass at G. Determine the normal reactions at each of the wheels at A and B if a force P = 50 N is applied to the handle. Neglect the mass and rolling resistance of the wheels. P

R17–3. The car has a mass of 1.50 Mg and a mass center at G. Determine the maximum acceleration it can have if power is supplied only to the rear wheels. Neglect the mass of the wheels in the calculation, and assume that the wheels that do not receive power are free to roll. Also, assume that slipping of the powered wheels occurs, where the coefficient of kinetic friction is mk = 0.3.

60

0.5 m

G

0.4 m

0.2 m A

B 0.3 m

B

1.6 m

0.4 m

0.2 m

G

1.3 m

A

Prob. R17–3

Prob. R17–1 17 R17–2. The two 1.5-kg rods EF and HI are fixed (welded) to the link AC at E. Determine the internal axial force Ex, shear force Ey, and moment ME, which the bar AC exerts on FE at E if at the instant u = 30°, link AB has an angular velocity v = 5 rad>s and an angular acceleration a = 8 rad>s2 as shown.

R17–4. A 20-kg roll of paper, originally at rest, is pinsupported at its ends to bracket AB. The roll rest against a wall for which the coefficient of kinetic friction at C is mC = 0.3. If a force of 40 N is applied uniformly to the end of the sheet, determine the initial angular acceleration of the roll and the tension in the bracket as the paper unwraps. For the calculation, treat the roll as a cylinder.

y

B

1m A

H

1.5 m

x

1m

F I

a  8 rad/s2 v  5 rad/s

13

12 5

E B C

C 1.5 m

a

A 120 mm

u  30 D

Prob. R17–2

60

P  40 N

Prob. R17–4

review proBleMs R17–5. At the instant shown, two forces act on the 15-kg slender rod which is pinned at O. Determine the magnitude of force F and the initial angular acceleration of the rod so that the horizontal reaction which the pin exerts on the rod is 25 N directed to the right.

471

R17–7. The spool and wire wrapped around its core have a mass of 20 kg and a centroidal radius of gyration kG = 250 mm. If the coefficient of kinetic friction at the ground is mB = 0.1, determine the angular acceleration of the spool when the 30-N # m couple moment is applied.

O

30 Nm 1.5 m

G

400 mm

100 N

200 mm 1.5 m

B F

Prob. R17–7

1m

17 Prob. R17–5

R17–6. The pendulum consists of a 15-kg sphere and a 5-kg slender rod. Compute the reaction at the pin O just after the cord AB is cut.

R17–8. Determine the backspin V which should be given to the 9-kg ball so that when its center is given an initial horizontal velocity vG = 6 m>s it stops spinning and translating at the same instant. The coefficient of kinetic friction is mA = 0.3.

v B

G

A O 0.3 m

0.6 m

Prob. R17–6

0.15 m

A

Prob. R17–8

vG  6 m/s

Chapter 18

(© Arinahabich/Fotolia) Roller coasters must be able to coast over loops and through turns, and have enough energy to do so safely. Accurate calculation of this energy must account for the size of the car as it moves along the track.

Planar Kinetics of a Rigid Body: Work and Energy Video Solutions are available for selected questions in this chapter.

Chapter ObjeCtives n

To develop formulations for the kinetic energy of a body, and define the various ways a force and couple do work.

n

To apply the principle of work and energy to solve rigid–body planar kinetic problems that involve force, velocity, and displacement.

n

To show how the conservation of energy can be used to solve rigid–body planar kinetic problems.

y

18.1

vi

Kinetic Energy

In this chapter we will apply work and energy methods to solve planar motion problems involving force, velocity, and displacement. But first it will be necessary to develop a means of obtaining the body’s kinetic energy when the body is subjected to translation, rotation about a fixed axis, or general plane motion. To do this we will consider the rigid body shown in Fig. 18–1, which is represented here by a slab moving in the inertial x–y reference plane. An arbitrary ith particle of the body, having a mass dm, is located a distance r from the arbitrary point P. If at the instant shown the particle has a velocity vi , then the particle’s kinetic energy is Ti = 12 dm v2i .

x i V r

y vP x

P

Fig. 18–1

474

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

and

energy

The kinetic energy of the entire body is determined by writing similar expressions for each particle of the body and integrating the results, i.e.,

y vi x

T =

i

1 dm v2i 2 Lm

V r

This equation may also be expressed in terms of the velocity of point P. If the body has an angular velocity V, then from Fig. 18–1 we have

y vP x

P

Fig. 18–1 (repeated)

vi = vP + vi>P = (vP)x i + (vP)y j + vk * (xi + yj) = [(vP)x - vy]i + [(vP)y + vx]j The square of the magnitude of vi is thus vi # vi = v2i = [(vP)x - vy]2 + [(vP)y + vx]2 = (vP)2x - 2(vP)xvy + v2y2 + (vP)2y + 2(vP)yvx + v2x2 = v2P - 2(vP)xvy + 2(vP)yvx + v2r2 Substituting this into the equation of kinetic energy yields

T = 18

1 1 a dm b v2P - (vP)xva y dm b + (vP)yva x dm b + v2 a r2 dm b 2 Lm 2 Lm Lm Lm The first integral on the right represents the entire mass m of the body. Since ym = 1 y dm and xm = 1 x dm, the second and third integrals locate the body’s center of mass G with respect to P. The last integral represents the body’s moment of inertia IP , computed about the z axis passing through point P. Thus, T = 12 mv2P - (vP)xvym + (vP)yvxm + 12 IPv2

(18–1)

As a special case, if point P coincides with the mass center G of the body, then y = x = 0, and therefore T = 12 mv2G + 12 IGv2

(18–2)

Both terms on the right side are always positive, since vG and v are squared. The first term represents the translational kinetic energy, referenced from the mass center, and the second term represents the body’s rotational kinetic energy about the mass center.

18.1

Translation. When a rigid body of mass m is subjected to either

v

rectilinear or curvilinear translation, Fig. 18–2, the kinetic energy due to rotation is zero, since V = 0. The kinetic energy of the body is therefore T = 12 mv2G

475

KinetiC energy

vG  v G

(18–3)

Rotation about a Fixed Axis.

When a rigid body rotates about a fixed axis passing through point O, Fig. 18–3, the body has both translational and rotational kinetic energy so that

Translation

Fig. 18–2

T =

1 2 2 mvG

+

1 2 2 IGv

(18–4) V

The body’s kinetic energy may also be formulated for this case by noting that vG = rGv, so that T = 12(IG + mr2G)v2. By the parallel–axis theorem, the terms inside the parentheses represent the moment of inertia IO of the body about an axis perpendicular to the plane of motion and passing through point O. Hence,* T =

1 2 2 IOv

vG G O

rG

(18–5)

From the derivation, this equation will give the same result as Eq. 18–4, since it accounts for both the translational and rotational kinetic energies of the body.

Rotation About a Fixed Axis

Fig. 18–3 18

General Plane Motion. When a rigid body is subjected to general plane motion, Fig. 18–4, it has an angular velocity V and its mass center has a velocity vG . Therefore, the kinetic energy is

V

vG

T =

1 2 2 mvG

+

1 2 2 IG v

(18–6)

This equation can also be expressed in terms of the body’s motion about its instantaneous center of zero velocity i.e., T = 12IICv2

G

(18–7)

where IIC is the moment of inertia of the body about its instantaneous center. The proof is similar to that of Eq. 18–5. (See Prob. 18–1.) *The similarity between this derivation and that of MO = IOa, should be noted. Also the same result can be obtained directly from Eq. 18–1 by selecting point P at O, realizing that vO = 0.

General Plane Motion

Fig. 18–4

476

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

and

energy

System of Bodies.

Because energy is a scalar quantity, the total kinetic energy for a system of connected rigid bodies is the sum of the kinetic energies of all its moving parts. Depending on the type of motion, the kinetic energy of each body is found by applying Eq. 18–2 or the alternative forms mentioned above.

The total kinetic energy of this soil compactor consists of the kinetic energy of the body or frame of the machine due to its translation, and the translational and rotational kinetic energies of the roller and the wheels due to their general plane motion. Here we exclude the additional kinetic energy developed by the moving parts of the engine and drive train.

18.2

The Work of a Force

Several types of forces are often encountered in planar kinetics problems involving a rigid body. The work of each of these forces has been presented in Sec. 14.1 and is listed below as a summary.

Work of a Variable Force. If an external force F acts on a body,

the work done by the force when the body moves along the path s, Fig. 18–5, is UF =

L

F # dr =

Ls

F cos u ds

(18–8)

Here u is the angle between the “tails” of the force and the differential displacement. The integration must account for the variation of the force’s direction and magnitude.

18 u

s

F

u F

Fig. 18–5 s

Fc

u Fc

Fc cos u

u Fc cos u

Fig. 18–6

Work of a Constant Force. If an external force Fc acts on a

body, Fig. 18–6, and maintains a constant magnitude Fc and constant direction u, while the body undergoes a translation s, then the above equation can be integrated, so that the work becomes UFc = (Fc cos u)s

(18–9)

18.2

477

the WorK of a forCe

Work of a Weight. The weight of a body does work only when the

body’s center of mass G undergoes a vertical displacement y. If this displacement is upward, Fig. 18–7, the work is negative, since the weight is opposite to the displacement.

G s

W y

UW = -W y

(18–10)

G W

Fig. 18–7

Likewise, if the displacement is downward (- y) the work becomes positive. In both cases the elevation change is considered to be small so that W, which is caused by gravitation, is constant.

Work of a Spring Force. If a linear elastic spring is attached to a body, the spring force Fs = ks acting on the body does work when the spring either stretches or compresses from s1 to a farther position s2 . In both cases the work will be negative since the displacement of the body is in the opposite direction to the force, Fig. 18–8. The work is Fs

k

Us = - 1 12 ks22 - 12 ks21 2

(18–11)

where  s2  7  s1  .

Unstretched position of spring, s  0

s1 s

Fig. 18–8

Forces That Do No Work.

There are some external forces that do no work when the body is displaced. These forces act either at fixed points on the body, or they have a direction perpendicular to their displacement. Examples include the reactions at a pin support about which a body rotates, the normal reaction acting on a body that moves along a fixed surface, and the weight of a body when the center of gravity of the body moves in a horizontal plane, Fig. 18–9. A frictional force Ff acting on a round body as it rolls without slipping over a rough surface also does no work.* This is because, during any instant of time dt, Ff acts at a point on the body which has zero velocity (instantaneous center, IC) and so the work done by the force on the point is zero. In other words, the point is not displaced in the direction of the force during this instant. Since Ff contacts successive points for only an instant, the work of Ff will be zero.

W V

r Ff IC

N *The work done by a frictional force when the body slips is discussed in Sec. 14.3.

18

s2

Fig. 18–9

478

Chapter 18

planar KinetiCs

of a

18.3 M

rigid Body: WorK

and

energy

The Work of a Couple Moment

Consider the body in Fig. 18–10a, which is subjected to a couple moment M = Fr. If the body undergoes a differential displacement, then the work done by the couple forces can be found by considering the displacement as the sum of a separate translation plus rotation. When the body translates, the work of each force is produced only by the component of displacement along the line of action of the forces dst , Fig. 18–10b. Clearly the “positive” work of one force cancels the “negative” work of the other. When the body undergoes a differential rotation du about the arbitrary point O, Fig. 18–10c, then each force undergoes a displacement dsu = (r>2) du in the direction of the force. Hence, the total work done is

u

(a)

F r F

dst

Translation (b)

r r dUM = Fa dub + Fa du b = (Fr) du 2 2 = M du

The work is positive when M and dU have the same sense of direction and negative if these vectors are in the opposite sense. When the body rotates in the plane through a finite angle u measured in radians, from u1 to u2 , the work of a couple moment is therefore

18 dsu

F du O du r r dsu 2 2 F

u2

UM =

Lu1

M du

(18–12)

Rotation (c)

Fig. 18–10

If the couple moment M has a constant magnitude, then

UM = M(u2 - u1)

(18–13)

479

the WorK of a Couple MoMent

18.3

ExamplE   18.1 The bar shown in Fig. 18–11a has a mass of 10 kg and is subjected to a couple moment of M = 50 N # m and a force of P = 80 N, which is always applied perpendicular to the end of the bar. Also, the spring has an unstretched length of 0.5 m and remains in the vertical position due to the roller guide at B. Determine the total work done by all the forces acting on the bar when it has rotated downward from u = 0 to u = 90.

B 0.75 m

M = 50 Nm A

k  30 N/m u

P  80 N

2m

Solution First the free-body diagram of the bar is drawn in order to account for all the forces that act on it, Fig. 18–11b.

1m (a)

Weight W. Since the weight 10(9.81) N = 98.1 N is displaced downward 1.5 m, the work is UW = 98.1 N(1.5 m) = 147.2 J Ay

Why is the work positive? Couple Moment M. The couple moment rotates through an angle of u = p>2 rad. Hence, UM = 50 N # m(p>2) = 78.5 J

u 50 Nm

Ax

Fs

P  80 N

1.5 m 98.1 N

spring Force Fs. When u = 0 the spring is stretched (0.75 m - 0.5 m) = 0.25 m, and when u = 90, the stretch is (2 m + 0.75 m) - 0.5 m = 2.25 m. Thus, Us = - 3 12(30 N>m)(2.25 m)2 - 12(30 N>m)(0.25 m)2 4 = -75.0 J

By inspection the spring does negative work on the bar since Fs acts in the opposite direction to displacement. This checks with the result. Force P. As the bar moves downward, the force is displaced through a distance of (p>2)(3 m) = 4.712 m. The work is positive. Why? UP = 80 N(4.712 m) = 377.0 J pin reactions. displaced.

Forces Ax and Ay do no work since they are not

total Work. The work of all the forces when the bar is displaced is thus U = 147.2 J + 78.5 J - 75.0 J + 377.0 J = 528 J

Ans.

0.5 m

1m

(b)

Fig. 18–11

18

480

Chapter 18

planar KinetiCs

of a

18.4

rigid Body: WorK

and

energy

Principle of Work and Energy

By applying the principle of work and energy developed in Sec. 14.2 to each of the particles of a rigid body and adding the results algebraically, since energy is a scalar, the principle of work and energy for a rigid body becomes T1 + U192 = T2

The counterweight on this bascule bridge does positive work as the bridge is lifted and thereby cancels the negative work done by the weight of the bridge.

(18–14)

This equation states that the body’s initial translational and rotational kinetic energy, plus the work done by all the external forces and couple moments acting on the body as the body moves from its initial to its final position, is equal to the body’s final translational and rotational kinetic energy. Note that the work of the body’s internal forces does not have to be considered. These forces occur in equal but opposite collinear pairs, so that when the body moves, the work of one force cancels that of its counterpart. Furthermore, since the body is rigid, no relative movement between these forces occurs, so that no internal work is done. When several rigid bodies are pin connected, connected by inextensible cables, or in mesh with one another, Eq. 18–14 can be applied to the entire system of connected bodies. In all these cases the internal forces, which hold the various members together, do no work and hence are eliminated from the analysis.

18

The work of the torque or moment developed by the driving gears on the motors is transformed into kinetic energy of rotation of the drum.

18.4

prinCiple of WorK and energy

481

procedure for analysis The principle of work and energy is used to solve kinetic problems that involve velocity, force, and displacement, since these terms are involved in the formulation. For application, it is suggested that the following procedure be used. Kinetic Energy (Kinematic Diagrams). • The kinetic energy of a body is made up of two parts. Kinetic energy of translation is referenced to the velocity of the mass center, T = 12 mv2G, and kinetic energy of rotation is determined using the moment of inertia of the body about the mass center, T = 12 IGv2. In the special case of rotation about a fixed axis (or rotation about the IC), these two kinetic energies are combined and can be expressed as T = 12 IOv2, where IO is the moment of inertia about the axis of rotation. • Kinematic diagrams for velocity may be useful for determining vG and v or for establishing a relationship between vG and v.* Work (Free–Body Diagram). • Draw a free–body diagram of the body when it is located at an intermediate point along the path in order to account for all the forces and couple moments which do work on the body as it moves along the path. • A force does work when it moves through a displacement in the direction of the force. • Forces that are functions of displacement must be integrated to obtain the work. Graphically, the work is equal to the area under the force–displacement curve. • The work of a weight is the product of its magnitude and the vertical displacement, UW = Wy. It is positive when the weight moves downwards. • The work of a spring is of the form Us = 12 ks2, where k is the spring stiffness and s is the stretch or compression of the spring. • The work of a couple is the product of the couple moment and the angle in radians through which it rotates, UM = Mu. • Since algebraic addition of the work terms is required, it is important that the proper sign of each term be specified. Specifically, work is positive when the force (couple moment) is in the same direction as its displacement (rotation); otherwise, it is negative. principle of Work and Energy. • Apply the principle of work and energy, T1 + U192 = T2 . Since this is a scalar equation, it can be used to solve for only one unknown when it is applied to a single rigid body. *A brief review of Secs. 16.5 to 16.7 may prove helpful when solving problems, since computations for kinetic energy require a kinematic analysis of velocity.

18

482

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

and

energy

ExamplE 18.2 The 30-kg disk shown in Fig. 18–12a is pin supported at its center. Determine the angle through which it must rotate to attain an angular velocity of 2 rad>s starting from rest. It is acted upon by a constant couple moment M = 5 N # m. The spring is orginally unstretched and its cord wraps around the rim of the disk. M  5 Nm

0.2 m O

k  10 N/m

(a)

Solution Kinetic energy. Since the disk rotates about a fixed axis, and it is initially at rest, then T1 = 0

18

T2 = 12 IOv22 =

294.3 N M  5 Nm

3

1 1 2 2 (30

kg)(0.2 m)2 4 (2 rad>s)2 = 1.2 J

Work (Free–body Diagram). As shown in Fig. 18–12b, the pin reactions Ox and Oy and the weight (294.3 N) do no work, since they are not displaced. The couple moment, having a constant magnitude, does positive work UM = Mu as the disk rotates through a clockwise angle of u rad, and the spring does negative work Us = - 12 ks2. principle of Work and energy.

O

5 T1 6 + 5 U1 - 2 6 = 5 T2 6

Ox

0.2 m Oy

Fs (b)

Fig. 18–12

5 0 6 + e (5 N # m)u -

5 T1 6 + e Mu - 12 ks2 f = 5 T2 6

1 (10 N>m)[u(0.2 m)]2 f = 5 1.2 J 6 2 - 0.2u2 + 5u - 1.2 = 0

Solving this quadratic equation for the smallest positive root, 180 u = 0.2423 rad = 0.2423 rad a b = 13.9 Ans. p rad

18.4

483

prinCiple of WorK and energy

ExamplE 18.3 The wheel shown in Fig. 18–13a weighs 20 kg and has a radius of gyration kG = 0.2 m about its mass center G. If it is subjected to a clockwise couple moment of 25 N # m and rolls from rest without slipping, determine its angular velocity after its center G moves 0.18 m. The spring has a stiffness k = 150 N>m and is initially unstretched when the couple moment is applied. Solution Kinetic energy (Kinematic Diagram). at rest,

k  150 N/m

G 0.25 m

(a)

A

The kinematic diagram of the wheel when it is in the final position is shown in Fig. 18–13b. The final kinetic energy is determined from

V2

0.5 m (vG)2

G

T2 = 12 IICv22

0.25 m

1 c 20 kg (0.2 m)2 + (20 kg)(0.25 m)2 d v22 2

IC (b)

T2 = 1.025 v22

Work (Free–body Diagram). As shown in Fig. 18–13c, only the spring force Fs and the couple moment do work. The normal force does not move along its line of action and the frictional force does no work, since the wheel does not slip as it rolls. The work of Fs is found using Us = - 12 ks2. Here the work is negative since Fs is in the opposite direction to displacement. Since the wheel does not slip when the center G moves 0.18 m, then the wheel rotates u = sG >rG>IC = 0.18 m>0.25 m = 0.72 rad, Fig. 18–13b. Hence, the spring stretches s = urA>IC = (0.72 rad)(0.5 m) = 0.36 m.

20 (9.81)N

Fs

25 Nm FB NB

principle of Work and energy. (c)

5 T1 6 + 5 U1 - 2 6 = 5 T2 6

5 0 6 + e 25 N # m(0.72 rad) -

25 Nm

Since the wheel is initially

T1 = 0

=

A

Fig. 18–13

5 T1 6 + 5 Mu - 12 ks2 6 = 5 T2 6

1 (150 N>m)(0.36 m)2 f = 5 1.025 v22 N # m 6 2 v2 = 2.84 rad>s b

Ans.

18

484

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

and

energy

ExamplE 18.4 The 700-kg pipe is equally suspended from the two tines of the fork lift shown in the photo. It is undergoing a swinging motion such that when u = 30 it is momentarily at rest. Determine the normal and frictional forces acting on each tine which are needed to support the pipe at the instant u = 0. Measurements of the pipe and the suspender are shown in Fig. 18–14a. Neglect the mass of the suspender and the thickness of the pipe. O u 0.4 m G 0.15 m

(a)

Fig. 18–14

Solution We must use the equations of motion to find the forces on the tines since these forces do no work. Before doing this, however, we will apply the principle of work and energy to determine the angular velocity of the pipe when u = 0. 18

Kinetic energy (Kinematic Diagram). at rest, then

Since the pipe is originally

T1 = 0 The final kinetic energy may be computed with reference to either the fixed point O or the center of mass G. For the calculation we will consider the pipe to be a thin ring so that IG = mr2. If point G is considered, we have T2 = 12 m(vG)22 + 12 IGv22 = 12(700 kg)[(0.4 m)v2]2 + 12[700 kg(0.15 m)2]v22 = 63.875v22 If point O is considered then the parallel-axis theorem must be used to determine IO . Hence, T2 = 12 IOv22 = 12[700 kg(0.15 m)2 + 700 kg(0.4 m)2]v22 = 63.875v22

18.4

prinCiple of WorK and energy

Work (Free-body Diagram). Fig. 18–14b. The normal and frictional forces on the tines do no work since they do not move as the pipe swings. The weight does positive work since the weight moves downward through a vertical distance y = 0.4 m - 0.4 cos 30 m = 0.05359 m.

485

NT FT

O u

principle of Work and energy.

0.4 m

5 T1 6 + 5 U1 - 2 6 = 5 T2 6

G

5 0 6 + 5 700(9.81) N(0.05359 m) 6 = 5 63.875v22 6

y

v2 = 2.400 rad>s

700 (9.81) N

equations of Motion. Referring to the free-body and kinetic diagrams shown in Fig. 18–14c, and using the result for v2, we have + Ft = m(aG)t; d

FT = (700 kg)(aG)t

+ c Fn = m(aG)n;

NT - 700(9.81) N = (700 kg)(2.400 rad>s)2(0.4 m)

c + MO = IOa;

0 = [(700 kg)(0.15 m)2 + (700 kg)(0.4 m)2]a

(b)

Since (aG)t = (0.4 m)a, then a = 0, (aG)t = 0 FT = 0 NT = 8.480 kN There are two tines used to support the load, therefore F =T = 0 8.480 kN NT= = = 4.24 kN 2

Ans. Ans.

NoTE: Due to the swinging motion the tines are subjected to a greater normal force than would be the case if the load were static, in which case NT= = 700(9.81) N>2 = 3.43 kN. NT FT

O

O 0.4 m

700 kg(aG)n =

0.4 m G

G 700 kg(aG)t

IGA

700 (9.81) N (c)

Fig. 18–14

18

486

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

and

energy

ExamplE 18.5

A

0.4 m u

G

0.4 m

The 10–kg rod shown in Fig. 18–15a is constrained so that its ends move along the grooved slots. The rod is initially at rest when u = 0. If the slider block at B is acted upon by a horizontal force P = 50 N, determine the angular velocity of the rod at the instant u = 45. Neglect friction and the mass of blocks A and B. Solution Why can the principle of work and energy be used to solve this problem? Kinetic energy (Kinematic Diagrams). Two kinematic diagrams of the rod, when it is in the initial position 1 and final position 2, are shown in Fig. 18–15b. When the rod is in position 1, T1 = 0 since (vG)1 = V1 = 0. In position 2 the angular velocity is V2 and the velocity of the mass center is (vG)2 . Hence, the kinetic energy is

P  50 N B (a)

T2 = 12 m(vG)22 + 12 IGv22 = 12(10 kg)(vG)22 + (vG)1  0

= 5(vG)22 + 0.2667(v2)2

G

v1  0 (vA)2 IC 1

A rG/IC

V2

18

45 (vB)2

G

45 0.4 m

Therefore,

(b)

NA

0.4 m 45 0.4 m (0.4 cos 45) m 98.1 N B (0.8 sin 45) m NB (c)

Fig. 18–15

= 0.4v2

T2 = 0.8v22 + 0.2667v22 = 1.0667v22 Of course, we can also determine this result using T2 =

2

A

kg)(0.8 m)2 4 v22

The two unknowns (vG)2 and v2 can be related from the instantaneous center of zero velocity for the rod. Fig. 18–15b. It is seen that as A moves downward with a velocity (vA)2 , B moves horizontally to the left with a velocity (vB)2 , Knowing these directions, the IC is located as shown in the figure. Hence, (vG)2 = rG>ICv2 = (0.4 tan 45 m)v2

(vG)2 0.4 m

B

50 N

3

1 1 2 12 (10

1 2 2 IICv2.

Work (Free–body Diagram). Fig. 18–15c. The normal forces NA and NB do no work as the rod is displaced. Why? The 98.1-N weight is displaced a vertical distance of y = (0.4 - 0.4 cos 45) m; whereas the 50-N force moves a horizontal distance of s = (0.8 sin 45) m. Both of these forces do positive work. Why? principle of Work and energy. 5 T1 6 + 5 U1 - 2 6 = 5 T2 6 5 T1 6 + 5 W y + Ps 6 = 5 T2 6 5 0 6 + 5 98.1 N(0.4 m - 0.4 cos 45 m) + 50 N(0.8 sin 45 m) 6 Solving for v2 gives

v2 = 6.11 rad>sb

= 5 1.0667v22 J 6

Ans.

487

prinCiple of WorK and energy

18.4

preliMinary prObleM P18–1.

Determine the kinetic energy of the 100-kg object.

3 rad/s

2 rad/s O

100 kg

30 100 kg 3m

(a) (d)

4m

2m O

2 rad/s

4 rad/s 100 kg 100 kg

(b) 2m

2 rad/s (e)

100 kg

3m 100 kg

2m

2m

4 rad/s No slipping (f)

(c)

Prob. P18–1

18

488

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

and

energy

FunDaMental prObleMs F18–1. The [email protected] wheel has a radius of gyration about its mass center O of kO = 400 mm. Determine its angular velocity after it has rotated 20 revolutions starting from rest. 0.6 m

P  50 N

F18–4. The [email protected] wheel is subjected to a force of 50 N. If the wheel starts from rest and rolls without slipping, determine its angular velocity after it has rotated 10 revolutions. The radius of gyration of the wheel about its mass center G is k G = 0.3 m. P  50 N

O 0.4 m

30 G

Prob. F18–1 F18–2. The uniform [email protected] slender rod is subjected to a couple moment of M = 150 N # m. If the rod is at rest when u = 0, determine its angular velocity when u = 90.

F18–5. If the uniform [email protected] slender rod starts from rest at the position shown, determine its angular velocity after it has rotated 4 revolutions. The forces remain perpendicular to the rod.

O M150 Nm

u

Prob. F18–4

1.5 m

30 N 0.5 m 0.5 m

0.5 m

1.5 m

18 O

20 Nm 20 N

Prob. F18–2 F18–3. The uniform [email protected] slender rod is at rest in the position shown when P = 600 N is applied. Determine the angular velocity of the rod when the rod reaches the vertical position. A

Prob. F18–5 F18–6. The [email protected] wheel has a radius of gyration about its center G of k G = 300 mm. When it is subjected to a couple moment of M = 50 N # m, it rolls without slipping. Determine the angular velocity of the wheel after its mass center G has traveled through a distance of sG = 20 m, starting from rest. 0.4 m

4m

5m M  50 Nm G P  600 N B

Prob. F18–3

Prob. F18–6

18.4

489

prinCiple of WorK and energy

prObleMs 18–1. At a given instant the body of mass m has an angular velocity V and its mass center has a velocity vG. Show that its kinetic energy can be represented as T = 12 IICv2, where IIC is the moment of inertia of the body determined about the instantaneous axis of zero velocity, located a distance rG>IC from the mass center as shown.

*18–4. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, and the wheel is rotated until the torque M = 25 N # m is developed, determine the maximum angular velocity of the wheel if it is released from rest. 18–5. The wheel is made from a 5-kg thin ring and two 2-kg slender rods. If the torsional spring attached to the wheel’s center has a stiffness k = 2 N # m>rad, so that the torque on the center of the wheel is M = (2u) N # m, where u is in radians, determine the maximum angular velocity of the wheel if it is rotated two revolutions and then released from rest.

IC rG/IC

G vG

v

0.5 m

Prob. 18–1 18–2. A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn since it is resting on the two rollers A and B of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the rollers and the mass of the cable. The radius of gyration of the reel about its center axis is kG = 0.42 m. 18–3. A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn without slipping on the two rollers A and B of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the cable. Each roller can be considered as an 18-kg cylinder, having a radius of 0.1 m. The radius of gyration of the reel about its center axis is kG = 0.42 m.

O M

18

Probs. 18–4/5 18–6. A force of P = 60 N is applied to the cable, which causes the 200-kg reel to turn since it is resting on the two rollers A and B of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the rollers and the mass of the cable. Assume the radius of gyration of the reel about its center axis remains constant at kO = 0.6 m.

P 30 P 250 mm G

O

0.75 m 1m

500 mm A

B 400 mm

Probs. 18–2/3

A

B 0.6 m

Prob. 18–6

490

Chapter 18

planar KinetiCs

of a

rigid Body: WorK

18–7. The wheel and the attached reel have a combined mass of 25 kg and a radius of gyration about their center of kA = 150 mm. If pulley B attached to the motor is subjected to a torque of M = 60(2 - e-0.1u) N # m, where u is in radians, determine the velocity of the 100-kg crate after it has moved upwards a distance of 1.5 m, starting from rest. Neglect the mass of pulley B.

and

energy

18–10. A motor supplies a constant torque M = 6 kN # m to the winding drum that operates the elevator. If the elevator has a mass of 900 kg, the counterweight C has a mass of 200  kg, and the winding drum has a mass of 600 kg and radius of gyration about its axis of k = 0.6 m, determine the speed of the elevator after it rises 5 m starting from rest. Neglect the mass of the pulleys.

*18–8. The wheel and the attached reel have a combined mass of 25 kg and a radius of gyration about their center of kA = 150 mm. If pulley B attached to the motor is subjected to a torque of M = 75 N # m, determine the velocity of the 100-kg crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

187.5 mm A

75 mm M

112.5 mm

B

M C

18

D

Probs. 18–7/8

0.8 m

Prob. 18–10

18–9. The pendulum consists of a 10-kg uniform disk and a 3-kg uniform slender rod. If it is released from rest in the position shown, determine its angular velocity when it rotates clockwise 90°.

18–11. The disk, which has a mass of 20 kg, is subjected to the couple moment of M = (2u + 4) N # m, where u is in radians. If it starts from rest, determine its angular velocity when it has made two revolutions.

300 mm A

B 0.8 m

M  30 N  m

O

D

2m

Prob. 18–9

Prob. 18–11

M

18.4 *18–12. The 10-kg uniform slender rod is suspended at rest when the force of F = 150 N is applied to its end. Determine the angular velocity of the rod when it has rotated 90° clockwise from the position shown. The force is always perpendicular to the rod.

prinCiple of WorK and energy

491

18–15. The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has rotated